11th NCERT/CBSE Binomial Theorem Exercise Miscellaneous
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Question (1)

Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.

Solution

(a+b)3, a1 = 729, a2 = 7290, a3 = 30375
a1 = nC0 an b0
729 = 1 ×an × 1
∴ 729 = an --- (1)

a2 = a1+1
7290 = nC1 an-1 b1
7290 = n an-1 b --- (2)

a3 = a2+1
30375 = nC2 an-2 b2
$30375 = \frac{{n\left( {n - 1} \right)}}{2}{a^{n - 2}}{b^2} - - - (3)$

(2) ÷ (1)
$\frac{{7290}}{{729}} = \frac{{n{a^{n - 1}}b}}{{{a^n}}}$
$10 = \frac{{nb}}{a}$ $ \therefore \frac{b}{a} = \frac{{10}}{n} - - - (4)$ (3) ÷ (2)
$\frac{{\require{cancel}\cancel{30375}^{25}}}{{\require{cancel}\cancel{7290}_6}} = \frac{{\cancel{n}\left( {n - 1} \right)}}{2}\frac{{{a^{n - 2}}{b^2}}}{{\cancel{n}{a^{n - 1}}b}}$
$\frac{{25}}{6} = \frac{{n - 1}}{2} \cdot \frac{b}{a}$
Replacing value of b/a from 4 we get
$\frac{{25}}{6} = \frac{{n - 1}}{\cancel{2}} \cdot \frac{{\cancel{10}^5}}{n}$
$\frac{{n - 1}}{n} = \frac{{\cancel{25}^5}}{6} \times \frac{1}{\cancel{5}} = \frac{5}{6}$
6n - 6 = 5n
n = 6
Replacing n = 6 in (1) we get
an = 729
a6 = 36
⇒ a = 3
Replacing value of n and a in (4)
$\frac{{10}}{n} = \frac{b}{a}$
$\frac{{10}}{6} = \frac{b}{3}$
⇒ b = 5
∴ a = 3, b = 5 and n = 6

Question (2)

Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.

Solution

(3+ax)9, n = 9, a = 3, b = ax
Let (r+1)th term be term of x3
∴ Tr+1 = nCr an-r br
Tr+1 = 9Cr (3)9-r (ax)r
Tr+1 = 9Cr (3)9-r ar xr
As it is term of x3
Power of x = 3
∴ r = 3 Replacing value we get
T4 = 9C3 36 a3 x3
∴ coeff. of x3 = 9C3 36 a3
Let (k+1)th term be term of x2
Tk+1 = nCk an-k bk Tk+1 =9Ck 39-k (ax)k
Tk+1 =9Ck 39-k ak xk
as it is term of x2
Power of x = 2
∴ k = 2
T3 = 9C2 (3)7 a2 x2
coeff. of x2 = 9C2 37 a2
Coefficient of x3 = coefficient of x2
9C3 36 a3 = 9C2 37 a2
$ \therefore \frac{{{a^3}}}{{{a^2}}} = \frac{{^9{C_2}{3^7}}}{{^9{C_3}{3^6}}}$
${a} = \frac{{\cancel{9} \times \cancel{8}}}{2} \times \frac{{3 \times \cancel{2}}}{{\cancel{9} \times \cancel{8} \times 7}} \times 3$
$a = \frac{9}{7}$

Question (3)

Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.

Solution

First expand (1 + 2x)6
=6C0 (2x)0 + 6C1(2x)1 + 6C2 (2x)2 +6C3(2x)3 + 6C4(2x)4 + 6C5 (2x)5 + 6C6 (2x)6
= 1+ 6(2x) + 15(4x2) + 20(8x3) +15(16x4) +6(32x5) +1(64x6)
= 1+12x +60x2 + 160x3 +240x4+192x5 + 64x6
Now (1-x)7
= 7C0 + 7C1(-x) + 7C2(-x)2 + 7C3(-x)3 + 7C4(-x)4 + 7C5(-x)5+ 7C6(-x)6 + 7C7(-x)7
=1-7x+21x2 -35x3+35x4-21x5+7x6-x7
coefficient of x5 in product
=1(-21) + 12(35) + 60(-35) +160(21) +240(-7) +192(1)
=-21+420-2100+3360-1680+192
=3972-3801
=171

Question (4)

If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.
[Hint: write an = (a – b + b)n and expand]

Solution

an = [(a-b) + b]n
Expanding by Bionomial we get
an = nC0(a-b)n b0 + nC1(a-b)n-1b + nC2 (a-b)n-2b2+ ... + nCn (a-b)0bn
an = (a-b)n + n(a-b)n-1b + nC2 (a-b)n-2b2 + ... + nC(a-b) bn-1 + bn
∴ an - bn = (a-b)n + n(a-b)n-1 b + nC2(a-b)n-2b2+ ...+ nCn-1(a-b)bn-1
an - bn = (a-b)[(a-b)n-1 + n(a-b)n-2b + nC2 (a-b)n-3 + ... + nCn-1 bn-1]
∴ So (a-b) is factor of an - bn

Question (5)

Evaluate ${\left( {\sqrt 3 + \sqrt 2 } \right)^6} - {\left( {\sqrt 3 - \sqrt 2 } \right)^6}$

Solution

Let √3 = a anda √2 = b, we get
(a+b)6
= 6C0 a6 + 6C1 a5b + 6C2 a4 b2 + 6C3 a3 b3 + 6C3 a2 b4+ 6C2 a b5 + 6C2 b6 ---(1)
=a6 +6a5b + 15a4b2 + 20 a3b3+15a2b4+6ab5+b6 ---(1)
Similarly (a-b)6
=a6 -6a5b + 15a4b2 - 20 a3b3+15a2b4-6ab5+b6 ---(2)
(1) - (2)
(a+b)6 - (a-b)6 =
= 12x5b + 40a3b3 + 12ab5
Replace a =√3 and b = √2 we get
\[{\left( {\sqrt 3 + \sqrt 2 } \right)^6} - {\left( {\sqrt 3 - \sqrt 2 } \right)^6}\] \[ = 12{\left( {\sqrt 3 } \right)^5}\left( {\sqrt 2 } \right) + 40{\left( {\sqrt 3 } \right)^3}{\left( {\sqrt 2 } \right)^3} + 12\left( {\sqrt 3 } \right){\left( {\sqrt 2 } \right)^5}\] \[ = 12\left( {9\sqrt 3 } \right)\left( {\sqrt 2 } \right) + 40\left( {3\sqrt 3 } \right)\left( {2\sqrt 2 } \right) + 12\sqrt 3 \left( {4\sqrt 2 } \right)\] \[ = 108\sqrt 6 + 240\sqrt 6 + 48\sqrt 6 \] \[ = 396\sqrt 6 \]

Question (6)

Find the value of ${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

Solution

\[{a^2} = x \; and \; \sqrt {{a^2} - 1} = y\] (x+y)4 + (x-y)4
(x+y)4
= 4C0 x4 + 4C1 x3 y + 4C2 x2 y2+ 4C3 x y3+ 4C4 x0 y4
= x4 +4x3y +6x2y2 +4xy3 + y4 ---(1)
Similarly (x-y)4
= x4 -4x3y +6x2y2 -4xy3 + y4 ---(2)
(1) + (2) we get
(x+y)4 + (x-y)4 = 2x4 +12x2y2+y4
Now replace value of x and y we get
${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$
$ = 2{\left( {{a^2}} \right)^4} + 12{\left( {{a^2}} \right)^2}{\left( {\sqrt {{a^2} - 1} } \right)^2} + 2{\left( {\sqrt {{a^2} - 1} } \right)^4}$
$ = 2{a^2} + 12{a^4}\left( {{a^2} - 1} \right) + 2{\left( {{a^2} - 1} \right)^2}$
$ = 2{a^8} + 12{a^6} - 12{a^4} + 2\left( {{a^4} - 2{a^2} + 1} \right)$
$ = 2{a^8} + 12{a^6} - 12{a^4} + 2{a^4} - 4{a^2} + 2$
$ = 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2$

Question (7)

Find an approximation of (0.99)5 using the first three terms of its expansion.

Solution

(0.99)5 = ( 1 - 0.01)5
${ = ^5}{C_0}{\left( { - 0.01} \right)^0}{ + ^5}{C_1}\left( { - 0.01} \right){ + ^5}{C_2}{\left( { - 0.01} \right)^2}$
=1 +5(-0.01) + 10(0.0001)
= 0.951

Question (8)

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${\left( {\sqrt[4]{2} + \frac{1}{{\sqrt[4]{3}}}} \right)^n}$ is $\sqrt 6 :1$

Solution

Let $\sqrt[4]{2} = a$ and $\frac{1}{{\sqrt[4]{3}}} = b$
(a+b)n
In the expansion there will be (n+1) terms
5th term from begining = 75
T5 = T4+1
T5 = nC4 an-4 b4 ---(1)
5th term of last = [(n+1) -5 +1]th from starting
=(n+1-5+1)th
=(n-3)th from starting
Tn-3 = Tn-4+1 ⇒ r = n - 4
= nCn-4 an-(n-4) bn-4
= nCn-4 a4 bn-4 ---(2)
As ratio of 5th term from starting anad 5th term from last is $\sqrt 6 :1$
$ \Rightarrow \frac{{{T_5}}}{{{T_{n - 3}}}} = \frac{{\sqrt 6 }}{1}$
$ \Rightarrow \frac{{^n{C_4}{a^{n - 4}}{b^4}}}{{^n{C_{n - 4}}{a^4}{b^{n - 4}}}} = \frac{{\sqrt 6 }}{1}$
As $^n{C_{n - 4}}{ = ^n}{C_4}$
$ \Rightarrow {\rm{ }}\frac{{^n{C_4}{a^{n - 4}}{b^4}}}{{^n{C_4}{a^4}{b^{n - 4}}}} = \frac{{\sqrt 6 }}{1}$
$\frac{{{a^{n - 8}}}}{{{b^{n - 8}}}} = \frac{{\sqrt 6 }}{1}$
Replace values of a anad b we get
${\left( {\frac{{\sqrt[4]{2}}}{{\frac{1}{{\sqrt[4]{3}}}}}} \right)^{n - 8}} = \frac{{\sqrt 6 }}{1}$
${\left( {\sqrt[4]{2} \cdot \sqrt[3]{3}} \right)^{n - 8}} = \sqrt 6 $
${\left( {\sqrt[4]{6}} \right)^{n - 8}} = \sqrt 6 $
\[{\left[ {{{\left( 6 \right)}^{\frac{1}{4}}}} \right]^{n - 8}} = {\left( 6 \right)^{\frac{1}{2}}}\] $ \Rightarrow \frac{{n - 8}}{4} = \frac{1}{2}$ 2(n-8) = 4
n - 8 =2
n = 10

Question (9)

Expand using Binomial Theorem ${\left( {1 + \frac{x}{2} - \frac{2}{x}} \right)^4},x \ne 0$

Solution

${\left( {1 + \frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = {\left[ {1 + \left( {\frac{x}{2} - \frac{1}{x}} \right)} \right]^4}$
${ = ^4}{C_0}{ + ^4}{C_1}\left( {\frac{x}{2} - \frac{2}{x}} \right){ + ^4}{C_2}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^2}{ + ^4}{C_3}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^3}{ + ^4}{C_4}{\left( {\frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = 1 + 4\left( {\frac{x}{2} - \frac{2}{x}} \right) + 6\left( {\frac{{{x^4}}}{4} - 2 + \frac{4}{{{x^2}}}} \right) + 4\left[ {\frac{{{x^3}}}{8} - \frac{8}{{{x^3}}} - 3\left( {\frac{x}{2} - \frac{2}{x}} \right)} \right] + {\left( {\frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = - 11 - 4x + \frac{{16}}{x} + \frac{{3{x^2}}}{2} + \frac{{24}}{{{x^2}}} + \frac{{{x^3}}}{2} - \frac{{32}}{{{x^3}}} + {\left( {\frac{x}{2} - \frac{2}{x}} \right)^4} - - - (i)$ Now we will solve ${\left( {\frac{x}{2} - \frac{2}{x}} \right)^4}$
${ = ^4}{C_0}{\left( {\frac{x}{2}} \right)^4}{ + ^4}{C_1}{\left( {\frac{x}{2}} \right)^3}\left( {\frac{{ - 2}}{x}} \right){ + ^4}{C_2}{\left( {\frac{x}{2}} \right)^2}{\left( {\frac{{ - 2}}{x}} \right)^2}{ + ^4}{C_3}\left( {\frac{x}{2}} \right){\left( {\frac{{ - 2}}{x}} \right)^3}{ + ^4}{C_4}{\left( {\frac{{ - 2}}{x}} \right)^4}$
$ = \frac{{{x^4}}}{{16}} + 4\frac{{{x^3}}}{8} \times \frac{{ - 2}}{x} + 6\frac{{{x^2}}}{4} \times \frac{4}{{{x^2}}} + 4\frac{x}{2} \times \frac{{ - 8}}{{{x^3}}} + \frac{{16}}{{{x^4}}}$ \[ = \frac{{{x^4}}}{{16}} - {x^2} + 6 - \frac{{16}}{{{x^2}}} + \frac{{16}}{{{x^4}}}\] Replace this value in (i) we get
${\left( {1 + \frac{x}{2} - \frac{2}{x}} \right)^4}$
$ = - 11 - 4x + \frac{{16}}{x} + \frac{{3{x^2}}}{2} + \frac{{24}}{{{x^2}}} + \frac{{{x^3}}}{2} - \frac{{32}}{{{x^3}}} + \frac{{{x^4}}}{{16}} - {x^2} + 6 - \frac{{16}}{{{x^2}}} + \frac{{16}}{{{x^4}}}$
$ = \frac{{{x^4}}}{{16}} + \frac{{{x^3}}}{2} + \frac{{{x^2}}}{2} - 4x - 5 + \frac{{16}}{x} + \frac{8}{{{x^2}}} - \frac{{32}}{{{x^3}}} + \frac{{16}}{{{x^4}}}$

Question (10)

Find the expansion of ${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$ using binomial theorem.

Solution

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3} = {\left[ {\left( {3{x^2} - 2ax} \right) + 3{a^2}} \right]^3}$
Using Binomial we get
$^3{C_0}{\left( {3{x^2} - 2ax} \right)^3}{ + ^3}{C_1}{\left( {3{x^2} - 2ax} \right)^2}\left( {3{a^2}} \right){ + ^3}{C_2}\left( {3{x^2} - 2ax} \right){\left( {3{a^2}} \right)^2}{ + ^3}{C_3}{\left( {3{a^2}} \right)^3}$
$ = 27{x^6} - 8{a^3}{x^3} - 18a{x^3}\left( {3{x^2} - 2ax} \right) + 3\left( {9{x^4} - 12a{x^3} + 4{a^2}{x^2}} \right)\left( {3{a^2}} \right) + 3\left( {3{x^2} - 2ax} \right)\left( {9{a^4}} \right) + 27{a^6}$
$ = 27{x^6} - 8{a^3}{x^3} - 54a{x^5} + 36{a^2}{x^4} + 81{a^2}{x^4} - 108{a^3}{x^3} + 36{a^4}{x^2} + 81{a^4}{x^2} - 54{a^5}x + 27{a^6}$
$ = 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$
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⇒8.2