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, a
1 = 729, a
2 = 7290, a
3 = 30375
a
1 =
nC
0 a
n b
0
729 = 1 ×a
n × 1
∴ 729 = a
n --- (1)
a
2 = a
1+1
7290 =
nC
1 a
n-1 b
1
7290 = n a
n-1 b --- (2)
a
3 = a
2+1
30375 =
nC
2 a
n-2 b
2
$30375 = \frac{{n\left( {n - 1} \right)}}{2}{a^{n - 2}}{b^2} - - - (3)$
(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
a
n = 729
a
6 = 3
6
⇒ 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 x
2 and x
3 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 x
3
∴ T
r+1 =
nC
r a
n-r b
r
T
r+1 =
9C
r (3)
9-r (ax)
r
T
r+1 =
9C
r (3)
9-r a
r x
r
As it is term of x
3
Power of x = 3
∴ r = 3
Replacing value we get
T
4 =
9C
3 3
6 a
3 x
3
∴ coeff. of x
3 =
9C
3 3
6 a
3
Let (k+1)
th term be term of x
2
T
k+1 =
nC
k a
n-k b
k
T
k+1 =
9C
k 3
9-k (ax)
k
T
k+1 =
9C
k 3
9-k a
k x
k
as it is term of x
2
Power of x = 2
∴ k = 2
T
3 =
9C
2 (3)
7 a
2 x
2
coeff. of x
2 =
9C
2 3
7 a
2
Coefficient of x
3 = coefficient of x
2
∴
9C
3 3
6 a
3 =
9C
2 3
7 a
2
$ \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 x
5 in the product (1 + 2x)
6 (1 – x)
7 using binomial theorem.
Solution
First expand (1 + 2x)
6
=
6C
0 (2x)
0 +
6C
1(2x)
1 +
6C
2 (2x)
2 +
6C
3(2x)
3 +
6C
4(2x)
4 +
6C
5 (2x)
5 +
6C
6 (2x)
6
= 1+ 6(2x) + 15(4x
2) + 20(8x
3) +15(16x
4) +6(32x
5) +1(64x
6)
= 1+12x +60x
2 + 160x
3 +240x
4+192x
5 + 64x
6
Now (1-x)
7
=
7C
0 +
7C
1(-x) +
7C
2(-x)
2 +
7C
3(-x)
3 +
7C
4(-x)
4 +
7C
5(-x)
5+
7C
6(-x)
6 +
7C
7(-x)
7
=1-7x+21x
2 -35x
3+35x
4-21x
5+7x
6-x
7
coefficient of x
5 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 a
n – b
n, whenever n is a positive integer.
[Hint: write a
n = (a – b + b)
n and expand]
Solution
a
n = [(a-b) + b]
n
Expanding by Bionomial we get
a
n =
nC
0(a-b)
n b
0 +
nC
1(a-b)
n-1b +
nC
2 (a-b)
n-2b
2+ ... +
nC
n (a-b)
0b
n
a
n = (a-b)
n + n(a-b)
n-1b +
nC
2 (a-b)
n-2b
2 + ... +
nC
(a-b) b
n-1 + b
n
∴ a
n - b
n = (a-b)
n + n(a-b)
n-1 b +
nC
2(a-b)
n-2b
2+ ...+
nC
n-1(a-b)b
n-1
a
n - b
n = (a-b)[(a-b)
n-1 + n(a-b)
n-2b +
nC
2 (a-b)
n-3 + ... +
nC
n-1 b
n-1]
∴ So (a-b) is factor of a
n - b
n
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
=
6C
0 a
6 +
6C
1 a
5b +
6C
2 a
4 b
2 +
6C
3 a
3 b
3 +
6C
3 a
2 b
4+
6C
2 a b
5
+
6C
2 b
6 ---(1)
=a
6 +6a
5b + 15a
4b
2 + 20 a
3b
3+15a
2b
4+6ab
5+b
6 ---(1)
Similarly (a-b)
6
=a
6 -6a
5b + 15a
4b
2 - 20 a
3b
3+15a
2b
4-6ab
5+b
6 ---(2)
(1) - (2)
(a+b)
6 - (a-b)
6 =
= 12x
5b + 40a
3b
3 + 12ab
5
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
=
4C
0 x
4 +
4C
1 x
3 y
+
4C
2 x
2 y
2+
4C
3 x y
3+
4C
4 x
0 y
4
= x
4 +4x
3y +6x
2y
2 +4xy
3 + y
4 ---(1)
Similarly (x-y)
4
= x
4 -4x
3y +6x
2y
2 -4xy
3 + y
4 ---(2)
(1) + (2) we get
(x+y)
4 + (x-y)
4 = 2x
4 +12x
2y
2+y
4
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
5
th term from begining = 75
T
5 = T
4+1
T
5 =
nC
4 a
n-4 b
4 ---(1)
5
th term of last = [(n+1) -5 +1]
th from starting
=(n+1-5+1)
th
=(n-3)
th from starting
T
n-3 = T
n-4+1 ⇒ r = n - 4
=
nC
n-4 a
n-(n-4) b
n-4
=
nC
n-4 a
4 b
n-4 ---(2)
As ratio of 5
th term from starting anad 5
th 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}$