11th NCERT/CBSE Binomial Theorem Exercise 8.1 Questions 14
Do or do not
There is no try

Expand each of the expressions in Exercise 1 to 5

Question (1)

(1- 2x)5

Solution

By using Binomial Theorem, the expression (1– 2x)5 can be expanded as \[{\left( {1 - 2x} \right)^5}\] \[{ = ^5}{C_0}{\left( 1 \right)^5}{ - ^5}{C_1}{\left( 1 \right)^4}{\left( {2x} \right)^1}{ + ^5}{C_2}{\left( 1 \right)^3}{\left( {2x} \right)^2}{ - ^5}{C_3}{\left( 1 \right)^2}{\left( {2x} \right)^3}{ + ^5}{C_4}{\left( 1 \right)^1}{\left( {2x} \right)^4}{ - ^5}{C_5}{\left( {2x} \right)^5}\] \[ = 1 - 5\left( {2x} \right) + 10\left( {4{x^2}} \right) - 10\left( {8{x^3}} \right) + 5\left( {16{x^4}} \right) - \left( {32{x^5}} \right)\] \[ = 1 - 10x + 40{x^2} - 80{x^3} + 80{x^4} - 32{x^5}\]

Question (2)

${\left( {\frac{2}{x} - \frac{x}{2}} \right)^2}$

Solution

By using Binomial Theorem, the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^2}$ can be expanded as \[{\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}{ = ^5}{C_0}{\left( {\frac{2}{x}} \right)^5}{ - ^5}{C_1}{\left( {\frac{2}{x}} \right)^4}{\left( {\frac{x}{2}} \right)^1}{ + ^5}{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{x}{2}} \right)^2}{ - ^5}{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{x}{2}} \right)^3}{ + ^5}{C_4}{\left( {\frac{2}{x}} \right)^1}{\left( {\frac{x}{2}} \right)^4}{ - ^5}{C_5}{\left( {\frac{x}{2}} \right)^5}\] \[ = \frac{{32}}{{{x^5}}} - 5\left( {\frac{{16}}{{{x^4}}}} \right)\left( {\frac{x}{2}} \right) + 10\left( {\frac{8}{{{x^3}}}} \right)\left( {\frac{{{x^2}}}{4}} \right) - 10\left( {\frac{4}{{{x^2}}}} \right)\left( {\frac{{{x^3}}}{8}} \right) + 5\left( {\frac{2}{x}} \right)\left( {\frac{{{x^4}}}{{16}}} \right) - \frac{{{x^5}}}{{32}}\] \[ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^3}}}{{32}}\]

Question (3)

(2x -3)6

Solution

By using Binomial Theorem, the expression (2x – 3)6 can be expanded as \[{\left( {2x - 3} \right)^6}{ = ^6}{C_0}{\left( {2x} \right)^6}{ - ^6}{C_1}{\left( {2x} \right)^5}\left( 3 \right){ + ^6}{C_2}{\left( {2x} \right)^4}{\left( 3 \right)^2}{ - ^6}{C_3}{\left( {2x} \right)^3}{\left( 3 \right)^3}{ + ^6}{C_4}{\left( {2x} \right)^4}{\left( 3 \right)^2}{ - ^6}{C_5}{\left( {2x} \right)^5}{\left( 3 \right)^1}{ + ^6}{C_6}{\left( {2x} \right)^6}\] \[ = 64{x^6} - 6\left( {32{x^5}} \right)\left( 3 \right) + 15\left( {16{x^4}} \right)\left( 9 \right) - 20\left( {8{x^3}} \right)\left( {27} \right) + 15\left( {4{x^2}} \right)\left( {81} \right) - 6\left( {2x} \right)\left( {243} \right) + 729\] \[ = 64{x^6} - 575{x^5} + 2160{x^4} - 4320{x^3} + 4860{x^2} - 2916x + 729\]

Question (4)

\[{\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}\]

Solution

By using Binomial Theorem, the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ can be expanded as \[{\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}{ = ^5}{C_0}{\left( {\frac{x}{3}} \right)^5}{ + ^5}{C_1}{\left( {\frac{x}{3}} \right)^4}{\left( {\frac{1}{x}} \right)^1}{ + ^5}{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2}{ + ^5}{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3}{ + ^5}{C_4}{\left( {\frac{x}{3}} \right)^1}{\left( {\frac{1}{x}} \right)^4}{ + ^5}{C_5}{\left( {\frac{1}{x}} \right)^5}\] \[ = \frac{{{x^5}}}{{243}} + 5\left( {\frac{{{x^4}}}{{81}}} \right)\left( {\frac{1}{x}} \right) + 10\left( {\frac{{{x^3}}}{{27}}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 10\left( {\frac{{{x^2}}}{9}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 5\left( {\frac{x}{3}} \right)\left( {\frac{1}{{{x^4}}}} \right) + \frac{1}{{{x^5}}}\] \[ = \frac{{{x^5}}}{{243}} + \frac{{5{x^3}}}{{81}} + \frac{{10x}}{{27}} + \frac{{10}}{{9x}} + \frac{5}{{3{x^2}}} + \frac{1}{{{x^5}}}\]

Question (5)

\[{\left( {x + \frac{1}{x}} \right)^5}\]

Solution

By using Binomial Theorem, the expression ${\left( {x + \frac{1}{x}} \right)^5}$ can be expanded as \[{\left( {x + \frac{1}{x}} \right)^6}{ = ^6}{C_0}{\left( x \right)^6}{ + ^6}{C_1}{\left( x \right)^5}\left( {\frac{1}{x}} \right){ + ^6}{C_2}{\left( x \right)^4}{\left( {\frac{1}{x}} \right)^2}{ + ^6}{C_3}{\left( x \right)^3}{\left( {\frac{1}{x}} \right)^3}{ + ^6}{C_4}{\left( x \right)^2}{\left( {\frac{1}{x}} \right)^4}{ + ^6}{C_5}{\left( x \right)^1}{\left( {\frac{1}{x}} \right)^5}{ + ^6}{C_6}{\left( {\frac{1}{x}} \right)^6}\] \[ = {x^6} + 6{\left( x \right)^5}\left( {\frac{1}{x}} \right) + 15{\left( x \right)^4}{\left( {\frac{1}{x}} \right)^2} + 20{\left( x \right)^3}{\left( {\frac{1}{x}} \right)^3} + 15{\left( x \right)^2}{\left( {\frac{1}{x}} \right)^4} + 6\left( x \right){\left( {\frac{1}{x}} \right)^5} + \frac{1}{{{x^6}}}\] \[ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}}\]

Question (6)

Using binomial theorem, evaluate: (96)3

Solution

96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.
It can be written that, 96 = 100 – 4 \[{\left( {96} \right)^3} = {\left( {100 - 4} \right)^3}\] \[{ = ^3}{C_0}{\left( {100} \right)^3}{ - ^3}{C_1}{\left( {100} \right)^2}\left( 4 \right){ + ^3}{C_2}\left( {100} \right){\left( 4 \right)^2}{ - ^3}{C_3}{\left( 4 \right)^3}\] \[ = {\left( {100} \right)^3} - 3{\left( {100} \right)^2}\left( 4 \right) + 3\left( {100} \right){\left( 4 \right)^2} - {\left( 4 \right)^3}\] \[ = 100000 - 120000 + 4800 - 64\] \[ = 884736\]

Question (7)

Using binomial theorem, evaluate: (102)5

Solution

102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 102 = 100 + 2 \[{\left( {102} \right)^5} = {\left( {100 + 2} \right)^5}\] \[{ = ^5}{C_0}{\left( {100} \right)^5}{ + ^5}{C_1}{\left( {100} \right)^4}\left( 2 \right){ + ^5}{C_2}{\left( {100} \right)^3}{\left( 2 \right)^2}{ + ^5}{C_3}{\left( {100} \right)^2}{\left( 2 \right)^3}{ + ^5}{C_4}{\left( {100} \right)^1}{\left( 2 \right)^4}{ + ^5}{C_5}{\left( 2 \right)^5}\] \[ = {\left( {100} \right)^5} + 5{\left( {100} \right)^4}\left( 2 \right) + 10{\left( {100} \right)^3}{\left( 2 \right)^2} + 10{\left( {100} \right)^2}{\left( 2 \right)^3} + 5\left( {100} \right){\left( 2 \right)^4} + {\left( 2 \right)^5}\] \[ = {\rm{10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32}}\] \[{\rm{ = 11040808032}}\]

Question (8)

Using binomial theorem, evaluate: (101)4

Solution

101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 101 = 100 + 1 \[{\left( {101} \right)^4} = {\left( {100 + 1} \right)^4}\] \[{ = ^4}{C_0}{\left( {100} \right)^4}{ + ^4}{C_1}{\left( {100} \right)^3}\left( 1 \right){ + ^4}{C_2}{\left( {100} \right)^2}{\left( 1 \right)^2}{ + ^4}{C_3}{\left( {100} \right)^1}{\left( 1 \right)^3}{ + ^4}{C_4}{\left( 1 \right)^4}\] \[ = 100000000 + 4000000 + 60000 + 400 + 1\] \[ = 104060401\]

Question (9)

Using binomial theorem, evaluate: (99)5

Solution

99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1 \[{\left( {99} \right)^5} = {\left( {100 - 1} \right)^5}\] \[{ = ^5}{C_0}{\left( {100} \right)^5}{ - ^5}{C_1}{\left( {100} \right)^4}\left( 1 \right){ + ^5}{C_2}{\left( {100} \right)^3}{\left( 1 \right)^2}{ - ^5}{C_3}{\left( {100} \right)^2}{\left( 1 \right)^3}{ + ^5}{C_4}{\left( {100} \right)^1}{\left( 1 \right)^4}{ + ^5}{C_5}{\left( 1 \right)^5}\] \[ = {\left( {100} \right)^5} - 5{\left( {100} \right)^4} + 10{\left( {100} \right)^3} - 10{\left( {100} \right)^2} + 5\left( {100} \right) - 1\] \[ = 10000000000 - 500000000 + 10000000 - 100000 + 500 - 1\] \[ = 9509900499\]

Question (10)

Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000

Solution

By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as \[{\left( {1.1} \right)^{10000}} = {\left( {1 + 0.1} \right)^{10000}}\] \[{ = ^{10000}}{C_0}{\left( 1 \right)^{10000}}{ + ^{10000}}{C_1}{\left( 1 \right)^{1000}}{\left( {.1} \right)^1} + \text{other positive terms}\] \[ = 1 + 10000\left( {0.1} \right) +\text{other positive terms} \] \[ > 1000\] Hemce (1.1)10000 > 1000

Question (11)

Find (a+b)4 - (a-b)4 hence, evaluate ${\left( {\sqrt 3 + \sqrt 2 } \right)^4} - {\left( {\sqrt 3 - \sqrt 2 } \right)^4}$

Solution

Using Binomial Theorem, the expressions, (a + b)4 and (a – b)4, can be expanded as \[{\left( {a + b} \right)^4}{ = ^4}{C_0}{a^4}{ + ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}{ + ^4}{C_3}{a^1}{b^3}{ + ^4}{C_4}{b^4}\] \[{\left( {a - b} \right)^4}{ = ^4}{C_0}{a^4}{ - ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}{ - ^4}{C_3}{a^1}{b^3}{ + ^4}{C_4}{b^4}\] \[{ = ^4}{C_0}{a^4}{ + ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}{ + ^4}{C_3}{a^1}{b^3}{ + ^4}{C_4}{b^4} - \left[ {^4{C_0}{a^4}{ - ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}{ - ^4}{C_3}{a^1}{b^3}{ + ^4}{C_4}{b^4}} \right]\] \[ = 2\left( {^4{C_1}{a^3}b{ + ^4}{C_3}a{b^3}} \right) = 2\left( {4{a^3}b + 4a{b^3}} \right)\] \[ = 8ab\left( {{a^2} + {b^2}} \right)\] By substituting $a = \sqrt 3 $   $b = \sqrt 2 $   we obtain \[{\left( {\sqrt 3 + \sqrt 2 } \right)^4} - {\left( {\sqrt 3 - \sqrt 2 } \right)^4} = 8\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} \right\}\] \[ = 8\left( {\sqrt 3 } \right)\left( {3 + 2} \right) = 40\sqrt 6 \]

Question (12)

Find (x+1)6 + (x-1)6. hence or otherwise evaluate ${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$

Solution

Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as \[{\left( {x + 1} \right)^6}{ = ^6}{C_0}{x^6}{ + ^6}{C_1}{x^5}{ + ^6}{C_2}{x^4}{ + ^6}{C_3}{x^3}{ + ^6}{C_4}{x^2}{ + ^6}{C_5}{x^1}{ + ^6}{C_6}{x^0}\] \[{\left( {x - 1} \right)^6}{ = ^6}{C_0}{x^6}{ - ^6}{C_1}{x^5}{ + ^6}{C_2}{x^4}{ - ^6}{C_3}{x^3}{ + ^6}{C_4}{x^2}{ - ^6}{C_5}{x^1}{ + ^6}{C_6}{x^0}\] \[{\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = 2\left[ {^6{C_0}{x^6}{ + ^6}{C_2}{x^4}{ + ^6}{C_4}{x^2}{ + ^6}{C_6}{x^0}} \right]\] \[{\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = 2\left[ {{x^6} + 15{x^4} + 15{x^2} + 1} \right]\] Substitute $x = \sqrt 2 $, we obtain \[{\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {{{\left( {\sqrt 2 } \right)}^6} + 15{{\left( {\sqrt 2 } \right)}^4} + 15{{\left( {\sqrt 2 } \right)}^2} + 1} \right]\] \[ = 2\left( {8 + 15 \times 4 + 15 \times 2 + 1} \right)\] \[ = 2\left( {8 + 60 + 30 + 1} \right)\] \[ = 2\left( {99} \right) = 198\]

Question (13)

Show that 9n+1 - 8n - 9 is divisible by 64, whenever n is positive integer

Solution

In order to show that 9n+1 -8n -9 is divisible by 64, it has to be proved that 9n+1 -8n -9 = 64k, where k is some natural number
By Binomial Theorem, \[{\left( {1 + a} \right)^m}{ = ^m}{C_0}{ + ^m}{C_1}a{ + ^m}{C_2}{a^2} + ...{ + ^m}{C_m}{a^m}\] For a = 8 and m = n + 1, we obtain \[{\left( {1 + 8} \right)^{n + 1}}{ = ^{n + 1}}{C_0}{ + ^{n + 1}}{C_1}8{ + ^{n + 1}}{C_2}{8^2} + ...{ + ^{n + 1}}{C_m}{8^{n - 1}}\] \[ \Rightarrow {9^{n + 1}} = 1 + \left( {n + 1} \right)\left( 8 \right) + {8^2}\left[ {^{n + 1}{C_2}{ + ^{n + 1}}{C_3}\left( 8 \right) + ...{ + ^{n + 1}}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right]\] \[ \Rightarrow {9^{n + 1}} = 1 + \left( {n + 1} \right)\left( 8 \right) + 64\left[ {^{n + 1}{C_2}{ + ^{n + 1}}{C_3}\left( 8 \right) + ...{ + ^{n + 1}}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right]\] \[ \Rightarrow {9^{n + 1}} = 1 + \left( {n + 1} \right)\left( 8 \right) + 64k\] Where $k{ = ^{n + 1}}{C_2}{ + ^{n + 1}}{C_3}\left( 8 \right) + ...{ + ^{n + 1}}{C_{n + 1}}{\left( 8 \right)^{n - 1}}$ is a natural number
Thus 9n+1 - 8n - 9 is divisible by 64, whenever n is positive integer

Question (14)

Prove that $\sum\limits_{r = 0}^n {{3^r}{\;^n}{C_r} = {4^n}} $

Solution

By Binomial Theorem, \[\sum\limits_{r = 0}^n {^n{C_r}} {a^{n - r}}{b^r} = {\left( {a + b} \right)^n}\] By putting b = 3 and a = 1 in the above equation, we obtain \[\sum\limits_{r = 0}^n {^n{C_r}} {\left( 1 \right)^{n - r}}{\left( 3 \right)^r} = {\left( {1 + 3} \right)^n}\] \[ \Rightarrow \sum\limits_{r = 0}^n {{3^r}} {\;^n}{C_r} = {4^n}\] Hnece, proved
#
⇒8.2