11th NCERT/CBSE Principle of mathematical induction Exercise 4.1 Questions 24
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Prove the following by using the principle of mathematical induction for all n∈ N:

Question (1)

$1 + 3 + {3^2} + .. + {3^{n - 1}} = \frac{{\left( {{3^n} - 1} \right)}}{2}$

Solution

I) Let us prove for n=1
\[LHS = {3^{1 - 1}} = {3^0} = 1\] \[RHS = \frac{{{3^1} - 1}}{2} = \frac{{3 - 1}}{2} = 1\] \[ = LHS\] ∴ P(n) is true for n = 1
II) Let us assume P(n) is true for n=k
\[1 + 3 + {3^2} + ... + {3^{k - 1}} = \frac{{{3^k} - 1}}{2} ---(i)\] III) Let us prove for n = k+ 1
We will prove that
\[1 + 3 + {3^2} + ... + {3^k} = \frac{{{3^{k + 1}} - 1}}{2} \] Consider
\[LHS = 1 + 3 + {3^2} + ... + {3^{\left( {k + 1} \right) - 1}}\] \[ = 1 + 3 + {3^2} + ... + {3^{k - 1}} + {3^{\left( {k + 1} \right) - 1}}\] \[ = \left( {1 + 3 + {3^2} + ... + {3^{k - 1}}} \right) + {3^k}\] Using (i)
\[ = \frac{{\left( {{3^k} - 1} \right)}}{2} + {3^k}\] \[ = \frac{{\left( {{3^k} - 1} \right) + 2 \cdot {3^k}}}{2}\] \[ = \frac{{\left( {1 + 2} \right){3^k} - 1}}{2}\] \[ = \frac{{3 \cdot {3^k} - 1}}{2}\] \[ = \frac{{{3^{k + 1}} - 1}}{2}\] Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural num,bers, i.e. n

Question (2)

${1^3} + {2^3} + {3^3} + ... + {n^3} = {\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)^2}$

Solution

\[{1^3} + {2^3} + {3^3} + ... + {n^3} = {\left( {\frac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] Let us prove for n=1
LHS = 13 = 1
\[RHS = {\left[ {\frac{{1\left( {1 + 1} \right)}}{2}} \right]^2} = 1 = LHS\] ∴ P(n) is true for n = 1
Let us assume P(n) is true for n=k
\[{1^2} + {2^2} + {3^2}........ + {k^2} = {\left[ {\frac{{k\left( {k + 1} \right)}}{2}} \right]^2} - - (i)\] Let us prove for n = k+1 \[{1^3} + {2^3} + {3^3} + ... + {\left( {k + 1} \right)^3} = {\left( {\frac{{\left( {k + 1} \right)\left( {k + 1 + 1} \right)}}{2}} \right)^2}\] \[ = {\left( {\frac{{\left( {k + 1} \right)\left( {k + 2} \right)}}{2}} \right)^2}\] Consider
\[LHS = {1^3} + {2^3} + .... + {\left( {k + 1} \right)^3}\] \[LHS = \left( {{1^3} + {2^3} + .... + {k^3}} \right) + {\left( {k + 1} \right)^3}\] Using I
\[ = {\left( {\frac{{k\left( {k + 1} \right)}}{2}} \right)^2} + {\left( {k + 1} \right)^3}\] \[ = \frac{{{k^2}{{\left( {k + 1} \right)}^2}}}{4} + {\left( {k + 1} \right)^3}\] \[ = \frac{{{k^2}{{\left( {k + 1} \right)}^2} + 4{{\left( {k + 1} \right)}^3}}}{4}\] \[ = \frac{{{{\left( {k + 1} \right)}^2}\left\{ {{k^2} + 4\left( {k + 1} \right)} \right\}}}{4}\] \[ = \frac{{{{\left( {k + 1} \right)}^2}\left\{ {{k^2} + 4k + 4} \right\}}}{4}\] \[ = \frac{{{{\left( {k + 1} \right)}^2}{{\left( {k + 2} \right)}^2}}}{4}\] Thus, P(k+1) is true whenever P(k) is true.
Hence, by the pinciple of mathematical induction, statement P(n) is ture for all natural numbers i.e. n

Question (3)

$1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + .. + \frac{1}{{\left( {1 + 2 + 3 + ...n} \right)}} = \frac{{2n}}{{\left( {n + 1} \right)}}$

Solution

\[1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + .. + \frac{1}{{\left( {1 + 2 + 3 + ...n} \right)}} = \frac{{2n}}{{\left( {n + 1} \right)}}\] Let us prove for n=1
\[LHS = \frac{1}{1} = 1\] \[RHS = \frac{{2\left( 1 \right)}}{{1 + 1}} = 1 = LHS\] ∴ P(n) is true for n =k
Let us assume it is true for n=k
\[1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + .. + \frac{1}{{\left( {1 + 2 + 3 + ... + k} \right)}} = \frac{{2k}}{{\left( {k + 1} \right)}} - - - (i)\] Let us prove for n=k+1
So we will prove
\[1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + .. + \frac{1}{{\left( {1 + 2 + 3 + ... + k + k + 1} \right)}} = \frac{{2\left( {k + 1} \right)}}{{\left( {k + 2} \right)}}\] \[LHS = 1 + \frac{1}{{\left( {1 + 2} \right)}} + \frac{1}{{\left( {1 + 2 + 3} \right)}} + .. + \frac{1}{{\left( {1 + 2 + 3 + ... + k + k + 1} \right)}}\] \[ = \left( {1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ... + \frac{1}{{1 + 2 + 3... + k}}} \right) + \frac{1}{{1 + 2 + 3 + ...k + \left( {k + 1} \right)}}\] Using I
USe formula \[1 + 2 + 3 + ... + n = \frac{{n\left( {n + 1} \right)}}{2}\]
\[ = \frac{{2k}}{{k + 1}} + \frac{1}{{\left( {\frac{{\left( {k + 1} \right)\left( {k + 1 + 1} \right)}}{2}} \right)}}\] \[ = \frac{{2k}}{{k + 1}} + \frac{2}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\] \[ = \frac{2}{{k + 1}}\left( {k + \frac{1}{{k + 2}}} \right)\] \[ = \frac{2}{{k + 1}}\left( {\frac{{k\left( {k + 2} \right) + 1}}{{k + 2}}} \right)\] \[ = \frac{2}{{k + 1}}\left( {\frac{{{k^2} + 2k + 1}}{{k + 2}}} \right)\] \[ = \frac{{2 \cdot {{\left( {k + 1} \right)}^2}}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\] \[ = \frac{{2 \cdot \left( {k + 1} \right)}}{{\left( {k + 2} \right)}}\] Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e. n

Question (4)

$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n\left( {n + 1} \right)\left( {n + 2} \right) = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)}}{4}$

Solution

Let us prove for n = 1
LHS = 1 (1+1) (1+2) = 6
\[RHS = \frac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)\left( {1 + 3} \right)}}{4} = 6 = LHS\] ∴ P(n) is true for n=1
Let us assume P(n) is true for n=k
\[ \therefore 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + n\left( {k + 1} \right)\left( {k + 2} \right) = \frac{{k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{4} - - - (i)\] Let us prove for n=k+1
So we will prove
\[1\cdot2\cdot3 + 2\cdot3\cdot4 + ... + \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right) = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\left( {k + 4} \right)}}{4}\] \[LHS = 1\cdot2\cdot3 + 2\cdot3\cdot4 + ... + \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\] \[ = \left\{ {1 \cdot 2 \cdot 3 + 2\cdot3\cdot4 + ... + k\left( {k + 1} \right)\left( {k + 2} \right)} \right\} + \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\] Using (i)
\[ = \frac{{k\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{4} + \left( {k + 1} \right)\left( {k + 2} \right)\left({k +3}\right) \] \[ = \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\left( {\frac{k}{4} + 1} \right)\] \[ = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)\left( {k + 4} \right)}}{4}\] Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement
P(n) is true for all natural numbers i.e. n

Question (5)

$1 \cdot 3 + 2 \cdot {3^2} + {3.3^3} + ... + n \cdot {3^n} = \frac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4}$

Solution

$1 \cdot 3 + 2 \cdot {3^2} + {3.3^3} + ... + n \cdot {3^n} = \frac{{\left( {2n - 1} \right){3^{n + 1}} + 3}}{4}$
Let us prove for n=1
\[LHS = 1 \cdot {3^1} = 3\] \[RHS = \frac{{\left[ {2\left( 1 \right) - 1} \right]{3^{1 + 1}} + 3}}{4}\] \[RHS = \frac{{9 + 3}}{4} = 3 = LHS\] ∴ P(n) is true for n=1
Let us assume P(n) is true for n = k
\[1 \cdot 3 + 2 \cdot {3^2} + 3 \cdot {3^3} + ... + k \cdot {3^k} = \frac{{\left( {2k - 1} \right){3^{k + 1}} + 3}}{4} - - - (i)\] Let us prove for n = k + 1
So we will prove that
\[1 \cdot 3 + 2 \cdot {3^2} + 3 \cdot {3^3} + ... + \left( {k + 1} \right) \cdot {3^{k + 1}} = \frac{{\left[ {2\left( {k + 1} \right) - 1} \right]{3^{k + 1 + 1}} + 3}}{4}\] \[ = \frac{{\left( {2k + 1} \right){3^{k + 2}} + 3}}{4}\] \[LHS = 1\cdot3 + 2\cdot{3^2} + ... + \left( {k + 1} \right){3^{k + 1}}\] \[ = \left( {1\cdot3 + 2\cdot{3^2} + 3 \cdot {3^2} + ... + k{3^k}} \right) + \left( {k + 1} \right){3^{k + 1}}\] From (i)
\[ = \frac{{\left( {2k - 1} \right){3^{k + 1}} + 3}}{4} + \left( {k + 1} \right){3^{k + 1}}\] \[ = \frac{{\left( {2k - 1} \right){3^{k + 1}} + 3 + 4\left( {k + 1} \right){3^{k + 1}}}}{4}\] \[ = \frac{{{3^{k + 1}}\left\{ {2k - 1 + 4\left( {k + 1} \right)} \right\} + 3}}{4}\] \[ = \frac{{{3^{k + 1}}\left\{ {6k + 3} \right\} + 3}}{4}\] \[ = \frac{{{3^{k + 1}} \cdot 3\left\{ {2k + 1} \right\} + 3}}{4}\] \[ = \frac{{{3^{\left( {k + 1} \right) + 1}}\left\{ {2k + 1} \right\} + 3}}{4}\] \[ = \frac{{{3^{k + 2}}\left( {2k + 1} \right) + 3}}{4} = RHS\] Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e. n

Question (6)

$1 \cdot 2 + 2 \cdot 3 + 3.4 + ... + n \cdot \left( {n + 1} \right) = \left[ {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}} \right]$

Solution

$1 \cdot 2 + 2 \cdot 3 + 3.4 + ... + n \cdot \left( {n + 1} \right) = \left[ {\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}} \right]$
Let us prove for n = 1
\[LHS = 1\left( {1 + 1} \right) = 2\] \[RHS = \frac{{1\left( {1 + 1} \right)\left( {1 + 2} \right)}}{3} = \frac{{1\left( 2 \right)\left( 3 \right)}}{3} = 2 = LHS\] Let us assume p(k) is true for n=k
\[\therefore 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + k\left( {k + 1} \right) = \frac{{k\left( {k + 1} \right)\left( {k + 2} \right)}}{3} - - - (i)\] Let us prove for n = k+1
so we will prove that
\[1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + \left( {k + 1} \right)\left( {k + 2} \right) = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{3}\] Left hand side is $1\cdot2 + 2\cdot3 + 3\cdot4 + ... + \left( {k + 1} \right)\left( {k + 2} \right)$ \[ = \left[ {1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + k\left( {k + 1} \right)} \right] + \left( {k + 1} \right) \cdot \left( {k + 2} \right)\] Using (i)
\[\frac{{\left( k \right)\left( {k + 1} \right)\left( {k + 2} \right)}}{3} + \left( {k + 1} \right)\left( {k + 2} \right)\] \[ = \left( {k + 1} \right) \cdot \left( {k + 2} \right)\left( {\frac{k}{3} + 1} \right)\] \[ = \frac{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}{3}\] = RHS
Thus P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural number i.e. n

Question (7)

$1 \cdot 3 + 3 \cdot 5 + 3.7 + ... + \left( {2n - 1} \right)\left( {2n + 1} \right) = \frac{{n\left( {4{n^2} + 6n - 1} \right)}}{3}$

Solution

Let us prove for n = 1
\[LHS = \left( {2 - 1} \right)\left( {2 + 1} \right) = 3\] \[RHS = \frac{{1\left( {4 + 6 - 1} \right)}}{3} = \frac{9}{3} = 3 = LHS\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k
\[1\cdot3 + 3\cdot5 + 3.7 + ... + \left( {2k - 1} \right)\left( {2k + 1} \right) = \frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3}\] Let us prove for n = k + 1, so we will prove
\[1\cdot3 + 3\cdot5 + 3.7 + ... + \left( {2k + 1} \right)\left( {2k + 3} \right) = \frac{{\left( {k + 1} \right)\left( {4{k^2} + 14k + 9} \right)}}{3}\] \[LHS = 1 \cdot 3 + 3 \cdot 5 + .... + \left( {2k + 1} \right)\left( {2k + 3} \right)\] \[1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + ... + \left( {2k - 1} \right)\left( {2k + 1} \right) + \left\{ {2\left( {k + 1} \right) - 1} \right\}\left\{ {2\left( {k + 1} \right) + 1} \right\}\] Using (i)
\[ = \frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3} + \left( {2k + 2 - 1} \right)\left( {2k + 2 + 1} \right)\] \[ = \frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3} + \left( {2k + 1} \right)\left( {2k + 3} \right)\] \[ = \frac{{k\left( {4{k^2} + 6k - 1} \right)}}{3} + \left( {4{k^2} + 8k + 3} \right)\] \[ = \frac{{k\left( {4{k^2} + 6k - 1} \right) + 3\left( {4{k^2} + 8k + 3} \right)}}{3}\] \[ = \frac{{4{k^3} + 6{k^2} - k + 12{k^2} + 24k + 9}}{3}\] \[ = \frac{{4{k^3} + 18{k^2} + 23k + 9}}{3}\] (K+ 1) is a factor of numaretor. \[ = \frac{{(k + 1)\left( {4{k^2} + 14k + 9} \right)}}{3}\] = RHS Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (8)

$1 \cdot 2 + 2 \cdot {2^2} + {3.2^3} + ... + 2 \cdot {2^n} = \left( {n - 1} \right){2^{n + 1}} + 2$

Solution

Let us prove for n = 1
\[LHS = 1 \cdot 2 = 2\] \[RHS = \left( {1 - 1} \right){2^{1 + 1}} + 2 = 0 + 2 = 2 = LHS\] ∴ P(n) is true for n=1
Let us assume p(n) is trure for n = 4 and therefor
\[1\cdot2 + 2\cdot{2^2} + {3.2^3} + ... + 2\cdot{2^k} = \left( {k - 1} \right){2^{k + 1}} + 2 - - - (i)\] Let us prove for n = k+1
So we will prove that
\[1 \cdot 2 + 2 \cdot {2^2} + 3 \cdot {2^3} + ... + \left( {k + 1} \right){2^{k + 1}} = k{2^{k + 2}} + 2\] \[LHS = 1 \cdot 2 + 2 \cdot {2^2} + ... + \left( {k + 1} \right){2^{k + 1}}\] \[\left\{ {1 \cdot 2 + 2 \cdot {2^2} + 3 \cdot {2^3} + ... + k \cdot {2^k}} \right\} + \left( {k + 1} \right) \cdot {2^{k + 1}}\] \[ = \left( {k - 1} \right){2^{k + 1}} + 2 + \left( {k + 1} \right){2^{k + 1}}\] \[ = {2^{k + 1}}\left\{ {\left( {k - 1} \right) + \left( {k + 1} \right)} \right\} + 2\] \[ = {2^{k + 1}} \cdot 2k + 2\] \[ = k\cdot{2^{\left( {k + 2} \right)}} + 2\] = RHS
Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (9)

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^n}}} = 1 - \frac{1}{{{2^n}}}$

Solution

Let us prove for n =1
\[LHS = \frac{1}{{{2^1}}} = \frac{1}{2}\] \[RHS = 1 - \frac{1}{{{2^n}}} = 1 - \frac{1}{2} = \frac{1}{2} = LHS\] ∴ P(n) is true for n=1
Let us assume it is true for n = k
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^k}}} = 1 - \frac{1}{{{2^k}}}\] Let us prove for n = k +1
So we will prove that
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^{k + 1}}}} = 1 - \frac{1}{{{2^{k + 1}}}}\] \[LHS = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^{k + 1}}}}\] \[\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{{{2^k}}}} \right) + \frac{1}{{{2^{k + 1}}}}\] Using (i)
\[ = \left( {1 - \frac{1}{{{2^k}}}} \right) + \frac{1}{{{2^{k + 1}}}}\] \[ = 1 - \frac{1}{{{2^k}}} + \frac{1}{{2 \cdot {2^k}}}\] \[ = 1 - \frac{1}{{{2^k}}}\left( {1 - \frac{1}{2}} \right)\] \[ = 1 - \frac{1}{{{2^k}}}\left( {\frac{1}{2}} \right)\] \[ = 1 - \frac{1}{{{2^{k + 1}}}}\] = RHS
Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (10)

$\frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ... + \frac{1}{{\left( {3n - 1} \right)\left( {3n + 2} \right)}} = \frac{n}{{\left( {6n + 4} \right)}}$

Solution

Let us prove for n=1
\[\frac{1}{{\left( {3 - 1} \right)\left( {3 + 2} \right)}} = \frac{1}{{10}}\] P(n) is true for n = 1
Let us assume it is true for n = k
\[\frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ... + \frac{1}{{\left( {3k - 1} \right)\left( {3k + 2} \right)}} = \frac{k}{{6k + 4}} - - - (i)\] Let us prove for n = k+1
We will prove \[\frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ... + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}} = \frac{{k + 1}}{{6k + 10}}\] \[LHS = \frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ... + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\] \[\frac{1}{{2 \cdot 5}} + \frac{1}{{5 \cdot 8}} + \frac{1}{{8 \cdot 11}} + ... + \frac{1}{{\left( {3k - 1} \right)\left( {3k + 2} \right)}} + \frac{1}{{\left\{ {3\left( {k + 1} \right) - 1} \right\}\left\{ {3\left( {k + 1} \right) + 2} \right\}}}\] Using (i)
\[ = \frac{k}{{6k + 4}} + \frac{1}{{\left( {3k + 3 - 1} \right)\left( {3k + 3 + 2} \right)}}\] \[ = \frac{k}{{6k + 4}} + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\] \[ = \frac{k}{{2\left( {3k + 2} \right)}} + \frac{1}{{\left( {3k + 2} \right)\left( {3k + 5} \right)}}\] \[ = \frac{1}{{\left( {3k + 2} \right)}}\left( {\frac{k}{2} + \frac{1}{{3k + 5}}} \right)\] \[ = \frac{1}{{\left( {3k + 2} \right)}}\left( {\frac{{k\left( {3k + 5} \right) + 2}}{{2\left( {3k + 5} \right)}}} \right)\] \[ = \frac{1}{{\left( {3k + 2} \right)}}\left( {\frac{{3{k^2} + 5k + 2}}{{2\left( {3k + 5} \right)}}} \right)\] \[ = \frac{1}{{\left( {3k + 2} \right)}}\left( {\frac{{\left( {3k + 2} \right)\left( {k + 1} \right)}}{{2\left( {3k + 5} \right)}}} \right)\] \[ = \frac{{\left( {k + 1} \right)}}{{6k + 10}}\] = RHS
Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (11)

$\frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{2 \cdot 3 \cdot 4}} + \frac{1}{{3 \cdot 4 \cdot 5}} + ... + \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{{n\left( {n + 3} \right)}}{{4\left( {n + 1} \right)\left( {n + 2} \right)}}$

Solution

Let us prove for n = 1
\[LHS = \frac{1}{{1\left( 1 \right)\left( 3 \right)}} = \frac{1}{6}\] \[RHS = \frac{{1\left( {1 + 3} \right)}}{{4\left( {1 + 1} \right)\left( {1 + 2} \right)}} = \frac{4}{{4\left( 2 \right)\left( 3 \right)}} = \frac{1}{6} = LHS\] Let us assume it is true for n = k
\[\frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{2 \cdot 3 \cdot 4}} + ... + \frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} - - - (i)\] Let us prove for n = k+1
We will prove that \[\frac{1}{{1\cdot2\cdot3}} + \frac{1}{{2\cdot3\cdot4}} + ... + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}} = \frac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}\] \[\left[ {\frac{1}{{1 \cdot 2 \cdot 3}} + \frac{1}{{2 \cdot 3 \cdot 4}} + \frac{1}{{3 \cdot 4 \cdot 5}} + ... + \frac{1}{{k\left( {k + 1} \right)\left( {k + 2} \right)}}} \right] + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\] Using (i)
\[ = \frac{{k\left( {k + 3} \right)}}{{4\left( {k + 1} \right)\left( {k + 2} \right)}} + \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k\left( {k + 3} \right)}}{4} + \frac{1}{{k + 3}}} \right\}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k{{\left( {k + 3} \right)}^2} + 4}}{{4\left( {k + 3} \right)}}} \right\}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k{{\left( {{k^2} + 6k + 9} \right)}^2} + 4}}{{4\left( {k + 3} \right)}}} \right\}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{{k^3} + 6{k^2} + 9k + 4}}{{4\left( {k + 3} \right)}}} \right\}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k\left( {{k^2} + 2k + 1} \right) + 4\left( {{k^2} + 2k + 1} \right)}}{{4\left( {k + 3} \right)}}} \right\}\] \[ = \frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}\left\{ {\frac{{k{{\left( {k + 1} \right)}^2} + 4{{\left( {k + 1} \right)}^2}}}{{4\left( {k + 3} \right)}}} \right\}\] \[ = \frac{{\left( {k + 1} \right)\left( {k + 4} \right)}}{{4\left( {k + 2} \right)\left( {k + 3} \right)}}\] = RHS
Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (12)

$a + ar + a{r^2} + ... + a{r^{n - 1}} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$

Solution

Let us prove for n=1
\[LHS = a{r^{1 - 1}} = a{r^0} = a\] \[RHS = \frac{{a\left( {r - 1} \right)}}{{r - 1}} = a = LHS\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k
\[a + ar + ... + a{r^{k - 1}} = \frac{{a\left( {{r^k} - 1} \right)}}{{r - 1}}\] Let us prove for n = k+1
We will prove that \[a + ar + ... + a{r^k} = \frac{{a\left( {{r^{k + 1}} - 1} \right)}}{{r - 1}}\] LHS = \[a + ar + a{r^2} + .....a{r^k}\] \[ = [a + ar + a{r^2} + ........a{r^{k - 1}}] + a{r^k}\] using (i) we get, \[ = \frac{{a\left( {{r^k} - 1} \right)}}{{r - 1}} + a{r^k}\] \[ = \frac{{a\left( {{r^k} - 1} \right) + a{r^k}\left( {r - 1} \right)}}{{r - 1}}\] \[ = \frac{{a\left[ {{r^k} - 1 + {r^{k + 1}} - {r^k}} \right]}}{{r - 1}}\] \[ = \frac{{a\left( {{r^{k + 1}} - 1} \right)}}{{r - 1}}\] = RHS.
Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (13)

$\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)\left( {1 + \frac{7}{9}} \right)....\left( {1 + \frac{{\left( {2n + 1} \right)}}{{{n^2}}}} \right) = {\left( {n + 1} \right)^2}$

Solution

Let us prove for n = 1 \[LHS = \left[ {1 + \frac{{2\left( 1 \right) + 1}}{{{1^2}}}} \right] = 1 + 3 = 4\] \[RHS = {\left( {n + 1} \right)^2} = {\left( {1 + 1} \right)^2} = 4 = LHS\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k
\[\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)....\left( {1 + \frac{{2k + 1}}{{{k^2}}}} \right) = {\left( {k + 1} \right)^2} ---(i)\] Let us prove for n = k +1
We will prove that
\[\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)....\left( {1 + \frac{{2k + 3}}{{{{\left( {k + 1} \right)}^2}}}} \right) = {\left( {k + 2} \right)^2}\] \[LHS = \left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)....\left( {1 + \frac{{2k + 3}}{{{{\left( {k + 1} \right)}^2}}}} \right)\] \[ = \left[ {\left( {1 + \frac{3}{1}} \right)\left( {1 + \frac{5}{4}} \right)....\left( {1 + \frac{{2k + 1}}{{{k^2}}}} \right)} \right]\left( {1 + \frac{{2k + 3}}{{{{\left( {k + 1} \right)}^2}}}} \right)\] using (i) we get, \[ = {\left( {k + 1} \right)^2}\left[ {\frac{{{{\left( {k + 1} \right)}^2} + 2k + 3}}{{{{\left( {k + 1} \right)}^2}}}} \right]\] \[ = \left[ {{k^2} + 2k + 1 + 2k + 3} \right]\] \[ = {k^2} + 4k + 4\] \[ = {\left( {k + 2} \right)^2} = RHS\] Thus, P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e. n

Question (14)

$\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right)....\left( {1 + \frac{1}{n}} \right) = \left( {n + 1} \right)$

Solution

Let us prove for n = 1
\[LHS = \left( {1 + \frac{1}{1}} \right) = 2\] \[RHS = n + 1 = 1 + 1 = 2 = LHS\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k
\[\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)......\left( {1 + \frac{1}{k}} \right) = k + 1\] Let us prove for n = k + 1
We will prove \[\left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)......\left( {1 + \frac{1}{{k + 1}}} \right) = k + 2\] \[LHS = \left( {1 + \frac{1}{1}} \right)\left( {1 + \frac{1}{2}} \right)......\left( {1 + \frac{1}{{k + 1}}} \right)\] Using 1 \[ = \left( {k + 1} \right)\left( {1 + \frac{1}{{k + 1}}} \right)\] \[ = \left( {k + 1} \right)\left( {\frac{{\left( {k + 1} \right) + 1}}{{k + 1}}} \right)\] \[ = \left( {k + 2} \right) \] Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (15)

${1^2} + {3^2} + {5^2} + ... + {\left( {2n - 1} \right)^2} = \frac{{n\left( {2n - 1} \right)\left( {2n + 1} \right)}}{3}$

Solution

Let us prove for n = 1
\[LHS = {\left[ {2\left( 1 \right) - 1} \right]^2} = 1\] \[RHS = \frac{{1\left( {2 - 1} \right)\left( {2 + 1} \right)}}{3} = 1 = RHS\] ∴ P(n) i true for n=1
Let us assume it is true for n = k \[{1^2} + {3^2} + ... + {\left( {2k - 1} \right)^2} = \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right)}}{3}\] Let us prove for n = k + 1 We will prove
\[{1^2} + {3^2} + ... + {\left( {2k + 1} \right)^2} = \frac{{\left( {k + 1} \right)\left( {2k + 1} \right)\left( {2k + 3} \right)}}{3}\] \[\left\{ {{1^2} + {3^2} + ... + {{\left( {2k + 1} \right)}^2}} \right\} + {\left\{ {2\left( {k + 1} \right) - 1} \right\}^2}\] Using i
\[ = \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right)}}{3} + {\left( {2k + 2 - 1} \right)^2}\] \[ = \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right)}}{3} + {\left( {2k + 1} \right)^2}\] \[ = \frac{{k\left( {2k - 1} \right)\left( {2k + 1} \right) + 3{{\left( {2k + 1} \right)}^2}}}{3}\] \[ = \frac{{\left( {2k + 1} \right)\left\{ {k\left( {2k - 1} \right) + 3\left( {2k + 1} \right)} \right\}}}{3}\] \[ = \frac{{\left( {2k + 1} \right)\left\{ {2{k^2} - k + 6k + 3} \right\}}}{3}\] \[ = \frac{{\left( {2k + 1} \right)\left\{ {2{k^2} + 5k + 3} \right\}}}{3}\] \[ = \frac{{\left( {2k - 1} \right)\left\{ {2{k^2} + 2k + 3k + 3} \right\}}}{3}\] \[ = \frac{{\left( {2k + 1} \right)\left\{ {2k\left( {k + 1} \right) + 3\left( {k + 1} \right)} \right\}}}{3}\] \[ = \frac{{\left( {2k + 1} \right)\left( {k + 1} \right)\left( {2k + 3} \right)}}{3}\] = RHS
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (16)

$\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + \frac{1}{{7 \cdot 10}} + ... + \frac{1}{{\left( {3n - 2} \right)\left( {3n + 1} \right)}} = \frac{n}{{\left( {3n + 1} \right)}}$

Solution

Let us prove for n = 1
\[LHS = \frac{1}{{\left( {3 - 2} \right)\left( {3 + 1} \right)}} = \frac{1}{4}\] \[RHS = \frac{1}{{3 + 1}} = \frac{1}{4} = LHS\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k \[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k - 2} \right)\left( {3k + 1} \right)}} = \frac{k}{{3k + 1}}\] Let us prove for n = k + 1
We will prove that
\[\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}} = \frac{{k + 1}}{{3k + 4}}\] \[LHS = \frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\] \[\left\{ {\frac{1}{{1 \cdot 4}} + \frac{1}{{4 \cdot 7}} + ... + \frac{1}{{\left( {3k - 2} \right)\left( {3k + 1} \right)}}} \right\} + \frac{1}{{\left\{ {3\left( {k + 1} \right) - 2} \right\}\left\{ {3\left( {k + 1} \right) + 1} \right\}}}\] using (i)
\[ = \frac{k}{{3k + 1}} + \frac{1}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\] \[ = \frac{1}{{3k + 1}}\left\{ {k + \frac{1}{{\left( {3k + 4} \right)}}} \right\}\] \[ = \frac{1}{{3k + 1}}\left\{ {\frac{{k\left( {3k + 4} \right) + 1}}{{\left( {3k + 4} \right)}}} \right\}\] \[ = \frac{1}{{3k + 1}}\left\{ {\frac{{3{k^2} + 4k + 1}}{{\left( {3k + 4} \right)}}} \right\}\] \[ = \frac{1}{{3k + 1}}\left\{ {\frac{{3{k^2} + 3k + k + 1}}{{\left( {3k + 4} \right)}}} \right\}\] \[ = \frac{{\left( {3k + 1} \right)\left( {k + 1} \right)}}{{\left( {3k + 1} \right)\left( {3k + 4} \right)}}\] \[ = \frac{{\left( {k + 1} \right)}}{{3k + 4}}\] = Rhs
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (17)

$\frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + \frac{1}{{7 \cdot 9}} + ... + \frac{1}{{\left( {2n + 1} \right)\left( {2n + 3} \right)}} = \frac{n}{{3\left( {2n + 3} \right)}}$

Solution

Let us prove for n = 1
\[LHS = \frac{1}{{3 \cdot 5}} = \frac{1}{{15}}\] \[RHS = \frac{1}{{3\left( {2 + 3} \right)}} = \frac{1}{{15}} = RHS\] ∴ P(n) is true for n = 1
Let us assume P(n) is true for n = k
\[\frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2k + 1} \right)\left( {2k + 3} \right)}} = \frac{k}{{3\left( {2k + 3} \right)}} - - - (i)\] Let us prove for n = k + 1
We will prove that
\[\frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2k + 3} \right)\left( {2k + 5} \right)}} = \frac{{k + 1}}{{3\left( {2k + 5} \right)}}\] \[LHS = \frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2k + 3} \right)\left( {2k + 5} \right)}}\] \[\left[ {\frac{1}{{3 \cdot 5}} + \frac{1}{{5 \cdot 7}} + ... + \frac{1}{{\left( {2k + 1} \right)\left( {2k + 3} \right)}}} \right] + \frac{1}{{\left\{ {2\left( {k + 1} \right) + 1} \right\}\left\{ {2\left( {k + 1} \right) + 3} \right\}}}\] Using (i) \[ = \frac{k}{{3\left( {2k + 3} \right)}} + \frac{1}{{\left( {2k + 3} \right)\left( {2k + 5} \right)}}\] \[ = \frac{1}{{\left( {2k + 3} \right)}}\left[ {\frac{k}{3} + \frac{1}{{\left( {2k + 5} \right)}}} \right]\] \[ = \frac{1}{{\left( {2k + 3} \right)}}\left[ {\frac{{k\left( {2k + 5} \right) + 3}}{{3\left( {2k + 5} \right)}}} \right]\] \[ = \frac{1}{{\left( {2k + 3} \right)}}\left[ {\frac{{2{k^2} + 5k + 3}}{{3\left( {2k + 5} \right)}}} \right]\] \[ = \frac{1}{{\left( {2k + 3} \right)}}\left[ {\frac{{2{k^2} + 2k + 3k + 3}}{{3\left( {2k + 5} \right)}}} \right]\] \[ = \frac{{\left( {k + 1} \right)\left( {2k + 3} \right)}}{{3\left( {2k + 3} \right)\left( {2k + 5} \right)}}\] \[ = \frac{{\left( {k + 1} \right)}}{{3\left\{ {2k + 5} \right\}}}\] = RHS
Thus P(k+1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (18)

$1 + 2 + 3 + ... + n < \frac{1}{8}{\left( {2n + 1} \right)^2}$

Solution

Let us prove for n = 1
\[LHS = 1\] \[RHS = \frac{1}{8}{\left( {2 + 1} \right)^2} = \frac{9}{8}\] \[1 < \frac{9}{8}\] ∴ P(n) is true for n = 1
Let us assume it is true for n = k
\[1 + 2 + 3 + ... + k < \frac{1}{8}{\left( {2k + 1} \right)^2}\] We will prove for n = k +1
We will prove \[1 + 2 + 3 + ... + k + 1 < \frac{1}{8}{\left( {2k + 3} \right)^2}\] \[LHS = 1 + 2 + 3.....\left( {k + 1} \right)\] using (i) \[1 + 2 + 3.....\left( {k + 1} \right) < \frac{1}{8}{\left( {2k + 1} \right)^2} + \left( {k + 1} \right)\] \[ < \frac{1}{8}\left\{ {{{\left( {2k +1 } \right)}^2} + 8\left( {k + 1} \right)} \right\}\] \[ < \frac{1}{8}\left\{ {4{k^2} + 4k + 1 + 8k + 8} \right\}\] \[ < \frac{1}{8}\left\{ {4{k^2} + 12k + 9} \right\}\] \[ < \frac{1}{8}{\left( {2k + 3} \right)^2}\] Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (19)

n(n+1)(n+5) is a multiple of 3

Solution

Let prove for n = 1
\[LHS = 1\left( {1 + 1} \right)\left( {1 + 5} \right) = 1\left( 2 \right)\left( 6 \right) = 12\] It is divisible by 3
∴ P(n) is divisible by 3 for n = 1
Let us assume it is true for n = k
∴ k(k+1)(k+5) is divisible by 3
\[k\left( {k + 1} \right)\left( {k + 5} \right) = 3m\] \[k + 1 = \frac{{3m}}{{k\left( {k + 5} \right)}}...(i)\] Let us prove for n = k+1
We will prove that (k+1)(k+2)(k+6) is divisible by 3
\[LHS = \left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 6} \right)\] Using (i) we get, \[ = \frac{{3m}}{{k\left( {k + 5} \right)}}\left( {k + 2} \right)\left( {k + 6} \right)\] \[ = 3\left[ {\frac{{m(k + 2)(k + 6)}}{{k(k + 5)}}} \right]\] We know that every real number a can be represented as a = bq + r where r = 0, 1, 2, ...b-1.
If b = 3 , then r = 0 , 1 , 2.
Now we represent k as k = 3q + r , r = 0 , 1 , 2
If r = 0, then k = 3q.
Replacing this value in LHS we get
\[LHS = 3\left[ {\frac{{m(k + 2)(k + 6)}}{{k(k + 5)}}} \right]\] \[ = 3\left[ {\frac{{m(3q + 2)(3q + 6)}}{{3q(3q + 5)}}} \right]\] \[ = 3\left[ {\frac{{m\left( {3q + 2} \right)3\left( {q + 2} \right)}}{{3q\left( {3q + 5} \right)}}} \right]\] \[ = 3\left[ {\frac{{m\left( {3q + 2} \right)\left( {q + 2} \right)}}{{q\left( {3q + 5} \right)}}} \right]\] So it is divisible by 3.
If r = 1, then k = 3q + 1
Replacing this value in LHS we get,
\[LHS = 3\left[ {\frac{{m(k + 2)(k + 6)}}{{k(k + 5)}}} \right]\] \[LHS = 3\left[ {\frac{{m(3q + 3)(3q + 7)}}{{\left( {3q + 1} \right)(3q + 6)}}} \right]\] \[3\left[ {\frac{{m3\left( {q + 1} \right)\left( {3q + 7} \right)}}{{\left( {3q + 1} \right)3\left( {q + 2} \right)}}} \right]\] \[ = 3\left[ {\frac{{m\left( {q + 1} \right)\left( {3q + 7} \right)}}{{\left( {3q + 1} \right)\left( {q + 2} \right)}}} \right]\] So it is divisible by 3, when k = 3q+1
If r = 2, then k = 3q + 2
Replacing this value in LHS we get,
\[LHS = 3\left[ {\frac{{m(k + 2)(k + 6)}}{{k(k + 5)}}} \right]\] \[LHS = 3\left[ {\frac{{m(3q + 4)(3q + 8)}}{{\left( {3q + 2} \right)(3q + 7)}}} \right]\] So it is divisible by 3 when k = 3q + 2
So for values of k, P(K + 1 ) is divisible by 3
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (20)

102n-1 + 1 is divisible by 11

Solution

102n-1+1 is divided by 11
Let us prove for n = 1
\[LHS = {10^1} + 1 = 11\] It is divsible by 11
Let us assume it is true for n = k
∴ 102k-1 + 1 is divisible by 11
102k-1 + 1 = 11m
102k-1 = 11m - 1 ---(i)
Let us prove for n = k +1
∴we will prove that 102k+1 is divisible by 11
\[LHS = {10^{2k + 1}} + 1\] \[ = {10^{2k - 1 + 2}} + 1\] \[ = {10^{2k - 1}} \times {10^2} + 1\] \[ = 100\left( {{{10}^{2k - 1}}} \right) + 1\] using (i) we get, \[ = 100\left( {11m - 1} \right) + 1\] \[ = 1100m - 100 + 1\] \[ = 1100m - 99\] \[ = 11\left( {100m - 9} \right)\] So it is divisible by 11
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (21)

x2n - y2n is divisible by x+y

Solution

Let us prove for n=1
x2 - y2 = (x + y) (x - y)
∴ It is divisible by x+y
∴ P(n) is true for n = 1
Let us assume it is true for n = k
∴ x2k - y2k is divisible by x + y
x2k - y2k = λ(x + y )
x2k = λ( x + y ) + y2k
Let us prove for n = k+1
∴ We will prove x2(k+1) - y2(k+1) is divisible by x + y
LHS = x2(k+1) - y2(y+1)
\[ = {x^{2k + 2}} - {y^{2k + 2}}\] \[ = {x^2} \cdot {x^{2k}} - {y^2} \cdot {y^{2k}}\] using (i) we get, \[ = {x^2}\left[ {\lambda \left( {x + y} \right) + {y^{2k}}} \right] - {y^2} \cdot {y^{2k}}\] \[ = \lambda {x^2}\left( {x + y} \right) + {x^2}{y^{2k}} - {y^2} \cdot {y^{2k}}\] \[ = \lambda {x^2}\left( {x + y} \right) + {y^{2k}}\left( {{x^2} - {y^2}} \right)\] \[ = \lambda {x^2}\left( {x + y} \right) + {y^{2k}}\left( {x - y} \right)\left( {x + y} \right)\] \[ = \left( {x + y} \right)\left[ {\lambda {x^2} + {y^{2k}}\left( {x - y} \right)} \right]\] Which has a factor of (x+y)
Thus, P(k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (22)

32n+2 - 8n - 9 is divisible by 8

Solution

Let us prove for n = 1
\[LHS = {3^{2 + 2}} - 8\left( 1 \right) - 9\] \[ = 81 - 8 - 9 = 64\] It is divisible by 8
∴ P(n) is true for n = 1
Let us assume it is true for n = k
${3^{2k + 2}} - 8k + 9$ is divisible by 8
\[{3^{2k + 2}} - 8k - 9 = 8m\] \[{3^{2k + 2}} = 8k + 8m + 9\] Let us prove for n = k + 1
We will prove
${3^{2k + 4}} - 8\left( {k + 1} \right) - 9$ is divisible 8
\[{3^{2k + 4}} - 8\left( {k + 1} \right) - 9\] \[ = {3^{2k + 2}} \cdot {3^2} - 8k - 8 - 9\] using (i) we get, \[ = 9\left( {8m + 8k + 9} \right) - 8k - 17\] \[ = 72m + 72k + 81 - 8k - 17\] \[ = 72m + 64k + 64\] \[ = 8\left( {9m + 8k + 8} \right)\] \[ = 8r\] Where $r = \left( {9m + 8k + 8} \right)$ is a natural number
Therefore ${3^{2\left( {k + 1} \right) + 2}} - 8\left( {k + 1} \right) - 9$ is divisible by 8
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (23)

41n-14n is a multiple of 27

Solution

Let us prove for n = 1
${41^1} - {14^1} = 27$ which is multiple of 27
∴ P(n) is true for n = 1
Let us assume it is true for n = k
${41^k} - {14^k}$ ismultiple of 27
\[{41^k} - {14^k} = 27m\] \[{41^k} = 27m - {14^k} - - - (i)\] Let us peove for n = k+1
We has to prove
${41^{k + 1}} - {14^{k + 1}}$ is multiple of 27 \[LHS = {41^{k + 1}} - {14^{k + 1}}\] \[ = {41^k}\cdot41 - {14^k}\cdot14\] \[ = 41\left( {27 + {{14}^k}} \right) - {14^k}\cdot14\] \[ = 41 \times 27 + 41 \times {14^k} - {14^k}\cdot14\] \[ = 41 \times 27 + {14^k}\left( {41 - 14} \right)\] \[ = 41 \times 27 + {14^k}(27)\] \[ = 27\left( {41 + {{14}^k}} \right)\] It is muliple of 27.
Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

Question (24)

(2n + 7) < (n + 3)3

Solution

Let us prove for n = 1
LHS = 2 + 7 = 9
RHS = ( 1 + 3 )2 = 42 = 16.
Since 9 <16
P(n) is true for n = 1 .
Let us assume that P(n) is true for n = k.
So (2n + 7 ) < ( k + 3 )2 ... (i)
Let us prove for n = k + 1
So we will prove that 2k + 9 < ( k + 4 )2
\[LHS = 2k + 9\] \[ = \left( {2k + 7} \right) + 2\] using (i) we get, \[ \prec {\left( {k + 3} \right)^2} + 2\] \[ = {k^2} + 6k + 9 + 2\] \[ = {k^2} + 8k + 16 - 2k - 5\] \[ = \left( {{k^2} + 8k + 16} \right) - \left( {2k + 5} \right)\] \[ = {\left( {k + 4} \right)^2} - \left( {2k + 5} \right)\] For every k ∈ N , 2k + 5 > 0
\[ \prec {\left( {k + 4} \right)^2}\] Thus, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.
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