# Class 11 Maths Ncert Solutions Ex 8.2

## Class 11 Maths Ncert Solutions Chapter 8 Ex 8.2

Formulas:

1. The general term in the expansion of (a + b)n :

Tr + 1 = nCr × (a)n – r × br

2. The middle term in the expansion of (a + b)n :

(a).  If n is even:

The middle term = (n2+1)thterm$\left ( \frac{n}{2}+1 \right )^{th}term$

(b).   If n is odd:

The middle term = (n+12)thtermand(n+12+1)thterm$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$

Q.1: Find the Coefficient of x6 in the expansion of (x + 2)9

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, on substituting n = 9, a = x and b = 2 in the above expression we will get:

Tr + 1 = 9Cr × (x)9 – r × 2r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x6, we will get:

i.e.         (x)9 – r = x6

(or)         9 – r = 6

Therefore, r = 3

Now, on substituting the value of r in equation (1) we will get:

T4 = 9C3 × (x)9 – 3 × 23

T4=9!3!6!×x6×8$\\\\\Rightarrow T_{4}=\frac{9!}{3!\;6!}\times x^{6}\times 8\\\\$ T4=9×8×7×6!3×2×1×6!×8x6=672x6$\\\\\Rightarrow T_{4}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 1\times 6!}\times 8\;x^{6}= 672\;x^{6}\\\\$

Therefore, the Coefficient of x6 in the expansion of (x + 2)9 = 672

Q.2: Find the Coefficient of a6 b8 in the expansion of (2a – 3b)14

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × yr

Therefore, on substituting n = 14, x = 2a   and y = (-3b) in the above expression we will get:

Tr + 1 = 14Cr × (2a)14 – r × (-3b)r

i.e.    Tr + 1 = [14Cr × (2)14 – r (-3)r ] (a)14 – r × (b)r  . . . . . . . . . .  (1)

Now, on comparing the coefficients of a and b in equation (1) with a6 b8, we will get:

i.e.         a6 b8 = a14 – r br

Therefore,        r = 8

Now, on substituting the value of ‘r’ in equation (1) we will get:

T9 = [ 14C8 × (2)14 – 8 (-3)8 ] (a)14 – 8 × (b)8

i.e.     T9 = [ 14C8 × (2)6 (-3)8 ] a6 b8

T9=14!8!6!×64×729×a6b8$\\\\\Rightarrow T_{9}=\frac{14!}{8!\;6!}\times 64\times 729\times a^{6}\;b^{8}\\\\$ T9=14×13×12×11×10×9×8!6×5×4×3×2×1×8!×64×729×a6b8$\\\\\Rightarrow T_{9}=\frac{14\times 13\times 12\times 11\times 10\times 9\times 8! }{6\times 5\times 4\times 3\times 2\times 1\times 8!}\times 64\times 729\times a^{6}\;b^{8}\\\\$ T9=3003×64×729×a6b8=140107968a6b8$\\\\\Rightarrow T_{9} = 3003\times 64\times 729\times a^{6}\;b^{8}=140107968\;\;a^{6}b^{8}\\\\$

Therefore, the Coefficient of a6 b8 in the expansion of (2a – 3b)14 = 140107968

Q.3: Write the general term in the expansion of (x3 – y2)5

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 5, a = x3 and b = y2 we will get:

Tr+1 = 5Cr × (x3)5 – r × (y2)r

Therefore, the general term in the expansion of (x3 – y2)5:

Tr+1 = 5Cr × (x)15 – 3 r × (y)2 r

Q.4: Write the general term in the expansion of (x4 – xy2)9

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 9, a = x4 and b = xy2 we will get:

Tr+1 = 9Cr × (x4)9 – r × (xy2)r

$\Rightarrow$ Tr+1 = 9Cr × (x)36 – 4 r × (x)2 r × (y)2 r

$\Rightarrow$ Tr+1 = 9Cr × (x)36 – 4 r + 2 r × (y)2 r

$\Rightarrow$ Tr+1 = 9Cr × (x)36 – 2 r × (y)2 r

Therefore, the general term in the expansion of (x4 – xy2)9:

Tr+1 = 9Cr × (x)36 – 2 r × (y)2r

Q.5: Find the 4th term in the expansion of (2x + 3y)10

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 3; for finding out the values of 4th Term (T4)

Therefore, T3 + 1 = nC3 × (a)n – 3 × b3

Now, on substituting n =10, a = 2x and b = 3y we will get:

$\\\Rightarrow$   T3 + 1 = 10C3 × (2x)10 – 3 × (3y)3

$\\\Rightarrow$   T4 = 10C3 × (2x)7 × (3y)3

T4=10!3!7!×128x7×27y3$\\\\\Rightarrow T_{4}=\frac{10!}{3!\; 7!}\times 128\;x^{7}\times 27\;y^{3}\\\\$ T4=10×9×8×7!3×2×1×7!×3456x7y3$\\\\\Rightarrow T_{4}=\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times 3456\;x^{7}y^{3}\\\\$ T4=120×3456x7y3=414720x7y3$\\\\\Rightarrow T_{4} = 120\times 3456\;x^{7}y^{3}=414720\;x^{7}y^{3}\\\\$

Therefore, 4th term in the expansion of (2x + 3y)10 = 414720 x7y3

Q.6 Find the 11th term in the expansion of (8x12x)15$\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}$

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Therefore, from the above equation the value of r should be 10 for finding out the values of 11th Term (T11)

Therefore, T10 + 1 = nC10 × (a)n – 10 × b10

Now, on substituting n =15, a = 8x and b = (12x)$\left (-\frac{1}{2\sqrt{x}} \right )$ we will get:

$\\\Rightarrow$   T10 + 1 = 15C10 × (8x)15 – 10 × (12x)10$\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\$

$\\\Rightarrow$   T11 = 15C10 × (8x)5 × (12x)10$\left (-\frac{1}{2\sqrt{x}} \right )^{10}\\$

T11=15!10!5!×8×8×8×8×8×x5×1210×(1x)10$\\\\\Rightarrow T_{11}=\frac{15!}{10!\; 5!}\times 8\times 8\times 8\times 8\times 8\times x^{5}\times \frac{1}{2^{10}}\times \left ( \frac{1}{\sqrt{x}} \right )^{10}\\\\$ T11=15×14×13×12×11×10!5×4×3×2×1×10!×32×x5×1x5$\\\\\\\Rightarrow T_{11}=\frac{15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times 1\times 10!}\times 32\times x^{5}\times \frac{1}{x^{5}}\\\\$ T11=3003×32=96096$\\\\\Rightarrow T_{11} = 3003\times 32\boldsymbol{=96096}\\\\$

Therefore, 11th term in the expansion of (8x12x)15$\left ( 8x-\frac{1}{2\sqrt{x}} \right )^{15}$= 96096

Q.7: In the expansion of (5x410)9$\left ( 5-\frac{x^{4}}{10} \right )^{9}$, Find the middle terms.

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = (n+12)thtermand(n+12+1)thterm$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$

Therefore, the middle terms in the expansion of (5x410)9$\left ( 5-\frac{x^{4}}{10} \right )^{9}$are:

(9+12)thtermand(9+12+1)thterm$\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term$

$\Rightarrow$ 5th term and 6th term

Now, 5th and 6th terms in the expansion of (5x410)9$\left ( 5-\frac{x^{4}}{10} \right )^{9}$ are:

T5 = T4 + 1 = 9C4 × (5)9 – 4 × (x410)4$\left ( -\frac{x^{4}}{10} \right )^{4}$

$\\\\\Rightarrow$   T5 = 9C4 × (5)5 × (x410)4$\left ( -\frac{x^{4}}{10} \right )^{4}\\\\$

T5=9!4!5!×(5)5×(x410)4$\\\\\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (5)^{5}\times \left ( -\frac{x^{4}}{10} \right )^{4}\\\\$ T5=9×8×7×6×5!4×3×2×1×5!×5×5×5×5×5×1104×x16$\\\\\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 5\times 5\times 5\times 5\times 5\times \frac{1}{10^{4}}\times x^{16}\\\\$ T5=126×516x16=3158x16$\\\\\Rightarrow T_{5} = 126\times \frac{5}{16}\;x^{16} \boldsymbol{=\frac{315}{8}\;x^{16}}\\\\$

Therefore, 5th term =3158x16$=\frac{315}{8}\;x^{16}$

Now, T6 = T5 + 1 = 9C5 × (5)9 – 5 × (x410)5$\left ( -\frac{x^{4}}{10} \right )^{5}$

$\\\\\Rightarrow$   T6 = 9C5 × (5)4 × (x410)4$\left ( -\frac{x^{4}}{10} \right )^{4}\\\\$

T6=9!5!4!×(5)4×(x410)5$\\\\\Rightarrow T_{6}=\frac{9!}{5!\; 4!}\times (5)^{4}\times \left ( -\frac{x^{4}}{10} \right )^{5}\\\\$ T6=9×8×7×6×5!5!×(4×3×2×1)×5×5×5×5×1105×x20$\\\\\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 5\times 5\times 5\times 5\times \frac{-1}{10^{5}}\times x^{20}\\\\$ T6=126×1320x20=63160x20$\\\\\Rightarrow T_{6} = -126\times \frac{1}{320}\;x^{20}\boldsymbol{=\frac{-63}{160}\;x^{20}}\\\\$

Therefore, 6th term =63160x20$=\frac{63}{160}\;x^{20}$

Therefore, 5th and 6th terms are the middle terms in the expansion of (5x410)9$\left ( 5-\frac{x^{4}}{10} \right )^{9}$

And also, 5th term =3158x16$=\frac{315}{8}\;x^{16}$ and 6th term =63160x20$=\frac{-63}{160}\;x^{20}$

Q.8: In the expansion of (x2+8y)10$\left ( \frac{x}{2}+8y \right )^{10}$, Find the middle term.

Sol.

Here, n = 10

When n is even, the middle term in the expansion of (a + b)n is given by:

(n2+1)thterm$\left ( \frac{n}{2}+1 \right )^{th}term$

Therefore, the middle term in the expansion of (x2+8y)10$\left ( \frac{x}{2}+8y \right )^{10}$ is:

(102+1)thterm$\Rightarrow \left ( \frac{10}{2}+1 \right )^{th}term$ = 6th term

Now, 6th term in the expansion of (x2+8y)10$\left ( \frac{x}{2}+8y \right )^{10}$ is:

T6 = T5 + 1 = 10C5 × (x2)105$\left ( \frac{x}{2} \right )^{10-5}$ × (8y)5

$\\\Rightarrow$   T6 = 10C5 × (x2)5$\left ( \frac{x}{2} \right )^{5}$ × (8y)5

T6=10!5!5!×(x2)5×(8y)5$\\\Rightarrow T_{6}=\frac{10!}{5!\; 5!}\times \left ( \frac{x}{2} \right )^{5}\times (8y)^{5}\\$ T6=10×9×8×7×6×5!5×4×3×2×1×5!×x532×8×8×8×8×8×y5$\\\Rightarrow T_{6}=\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{x^{5}}{32}\times 8\times 8\times 8\times 8\times 8\times y^{5}\\$ T6=252×1024x5y5=258048x5y5$\\\Rightarrow T_{6} = 252\times 1024\;x^{5}y^{5}\boldsymbol{=258048\;x^{5}y^{5}}\\$

Therefore, 6th term = 258048 x5 y5

Hence, the middle term in the expansion of (x2+8y)10$\left ( \frac{x}{2}+8y \right )^{10}$ = 258048 x5 y5

Q.9: Prove that the coefficients of pand pb are equal in the expansion of (1 + p)a + b

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

Now, on substituting n = (a + b), x =1 and y = p, we will get:

Tr + 1 = (a + b)Cr × (1)(a + b) – r × (p)r

$\Rightarrow$ Tr + 1 = (a + b)Cr × (p)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of p in equation (1) with pa, we will get r = a

Therefore, from equation (1):

T a + 1 = (a + b)Ca × (p)a

Ta+1=(a+b)!a!(a+ba)!×pa$\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;(a+b-a)!}\times p^{a}\\$ Ta+1=(a+b)!a!b!×pa$\\\Rightarrow T_{a+1}=\frac{(a+b)!}{a!\;b!}\times p^{a}\\$

Therefore, the coefficient of (p)a =  (a+b)!a!b!$\frac{(a+b)!}{a!\;b!}$ . . . . . . . . . . (2)

Now, on comparing the coefficient of p in equation (1) with pb, we will get r = b

Therefore, from equation (1):

T b + 1 = (a + b)Cb × (p)b

Tb+1=(a+b)!b!(a+bb)!×pb$\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;(a+b-b)!}\times p^{b}\\\\$ Tb+1=(a+b)!b!a!×pb$\\\\\Rightarrow T_{b+1}=\frac{(a+b)!}{b!\;a!}\times p^{b}\\\\$

Therefore, the coefficient of (p)b(a+b)!b!a!$\frac{(a+b)!}{b!\;a!}$ . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3) we conclude that the coefficients of pand pb are equal in the expansion of (1 + p)a + b.

Hence, Proved

Q.10: The coefficients of the (k + 1)th, kth and (k – 1)th terms in the expansion of (x + 1)n are in the ratio 5 : 3 : 1. Find the values of n and k.

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now on substituting a = x and b = 1 in the above equation:

Tr + 1 = nCr × (x)n – r × (1)r

Now, (k + 1)th term in the expansion of (x + 1)n :

T(k) + 1 = nCk × (x)n – k × 1(k )

i.e.  T(k + 1) = nCk × (x)n – k

Tk+1=n!k!(nk)!×(x)nk$\\\Rightarrow T_{k+1} = \frac{n!}{k!\;(n-k)!}\times (x)^{n-k}\\$

Therefore the coefficient of T(k + 1)th term = n!k!(nk)!$\frac{n!}{k!\;(n-k)!}\\\\$

Now, (k)th term in the expansion of (x + 1)n :

T(k – 1) + 1 = nCk – 1 × (x)n – (k – 1) × 1(k – 1)

i.e.  Tk = nCk – 1 × (x)n – k + 1

Tk=n!(k1)!(nk+1)!×(x)nk+1$\\\\\Rightarrow T_{k} = \frac{n!}{(k-1)!\;(n-k+1)!}\times (x)^{n-k+1}\\\\$

Therefore the coefficient of  T(k)th term = n!(k1)!(nk+1)!$\frac{n!}{(k-1)!\;(n-k+1)!}$

And, (k – 1)th term in the expansion of (x + 1)n :

T(k – 2 )+ 1 = nCk – 2 × (x)n – (k – 2) × 1(k – 2)

i.e.  Tk – 1 = nCk – 2 × (x)n – k + 2

Tk1=n!(k2)!(nk+2)!×(x)nk+2$\\\\\Rightarrow T_{k-1} = \frac{n!}{(k-2)!\;(n-k+2)!}\times (x)^{n-k+2}\\\\$

Therefore the coefficient of  T(k – 1)th term = n!(k2)!(nk+2)!$\frac{n!}{(k-2)!\;(n-k+2)!}$

Now, according to the given conditions: Coefficients of T(k + 1), Tk and Tk – 1 are in the ratio of 5 : 3 : 1

Therefore, Tk1Tk=13$\frac{T_{k-1}}{T_{k}}=\frac{1}{3}\\$

[n!(k2)!(nk+2)!]÷[n!(k1)!(nk+1)!]=13$\\\\\Rightarrow \left [ \frac{n!}{(k-2)!\;(n-k+2)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{1}{3}\\\\$ (k1)!(nk+1)!(k2)!(nk+2)!=13$\\\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{(k-2)!\;(n-k+2)!}=\frac{1}{3}\\\\$ (k1)(k2)!(nk+1)!(k2)!(nk+2)(nk+1)!=13$\\\\\Rightarrow \frac{(k-1)(k-2)!\;(n-k+1)!}{(k-2)!\;(n-k+2)(n-k+1)!}=\frac{1}{3}\\\\$ k1nk+2=13$\\\Rightarrow \frac{k-1}{n-k+2}=\frac{1}{3}\\$ 3k3=nk+2$\\\Rightarrow 3k-3=n-k+2\\$

(or)              n – 4k + 5 = 0 . . . . . . . . . . . . . . . .  (1)

And,        Tk+1Tk=53$\frac{T_{k+1}}{T_{k}}=\frac{5}{3}\\$

[n!k!(nk)!]÷[n!(k1)!(nk+1)!]=53$\\\Rightarrow \left [ \frac{n!}{k!\;(n-k)!} \right ]\div \left [ \frac{n!}{(k-1)!\;(n-k+1)!} \right ]=\frac{5}{3}\\$ (k1)!(nk+1)!k!(nk)!=53$\\\Rightarrow \frac{(k-1)!\;(n-k+1)!}{k!\;(n-k)!}=\frac{5}{3}\\$ (k1)![(nk+1)(nk)!][k(k1)!](nk)!=53$\\\Rightarrow \frac{(k-1)!\;[(n-k+1)\;(n-k)!]}{[k\;(k-1)!]\;(n-k)!}=\frac{5}{3}\\$ nk+1k=53$\\\Rightarrow \frac{n-k+1}{k}=\frac{5}{3}\\$ 3n3k+3=5k$\\\Rightarrow 3n-3k+3=5k\\$

(or) 3n – 8k + 3 = 0 . . . . . . . . . . . . . . . . . . (2)

On multiplying equation (1) by 3 and subtracting it to equation (2) we will get:

(3n – 8k + 3) – 3(n – 4k + 5) = 0

$\\\Rightarrow$ 3n – 8k + 3 – 3n + 12k – 15 = 0

$\\\Rightarrow$ 4k = 12

Therefore, k = 3

Now on substituting the value of k in equation (2) we will get:

3n – 8(3) + 3 = 0

Therefore, n = 7

Q.11: Prove that the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

Sol.

Since, The general term in the expansion of (x + y)n is given by: Tr + 1 = nCr × (x)n – r × (y)r

For (1 + 3a)2p:

On putting n = 2p, x =1 and y = 3a, we will get:

Tr + 1 = (2p)Cr × (1)(3p) – r × (3a)r

$\\\\\Rightarrow$ Tr + 1 = (2p)Cr × (3a)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of a in equation (1) with ap, we will get r = p

Therefore, from equation (1):

T p + 1 = (2p)Cp × (3a)p

Tp+1=(2p)!p!(2pp)!×(3a)p$\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;(2p-p)!}\times (3a)^{p}\\\\$

Tp+1=(2p)!p!p!×3p×(a)p$\\\Rightarrow T_{p+1}=\frac{(2p)!}{p!\;p!}\times 3^{p}\times (a)^{p}\\\\$

Therefore, the coefficient of ap =(2p)!p!p!×3p$=\frac{(2p)!}{p!\;p!}\times 3^{p}$ . . . . . . . . . . . . . . . . (2)

Now, for (1 + 3a)2p – 1 :

On substituting n = (2p – 1), x =1 and y = 3a, we will get:

Tr + 1 = (2p – 1)Cr × (1)(2p – 1) – r × (3a)r

$\Rightarrow$ Tr + 1 = (2p – 1)Cr × (3a)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of ‘a’ in equation (1) with ‘ap ’, we will get r = p

Therefore, from equation (1):

T p + 1 = (2p – 1)Cp × (3a)p

Tp+1=(2p1)!p!(2p1p)!×(3a)p$\\\\\Rightarrow T_{p+1}=\frac{(2p-1)!}{p!\;(2p-1-p)!}\times (3a)^{p}\\\\$ Tp+1=(2p1)!p!(p1)!×(3)p×(a)p$\\\\\Rightarrow T_{p+1}= \frac{(2p-1)!}{p!\;(p-1)!}\times (3)^{p}\times (a)^{p}\\\\$ Tp+1=(2p)!2pp!×p!p×(3)p.(a)p=(2p)!2(p!)2×3p.ap$\\\\\Rightarrow T_{p+1}=\frac{\frac{(2p)!}{2p}}{p!\times \frac{p!}{p}}\times (3)^{p}.(a)^{p}=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}.a^{p}\\$

Therefore, the coefficient of ap = =(2p)!2(p!)2×3p$=\frac{(2p)!}{2(p!)^{2}}\times 3^{p}$ . . . . . . . . . . (3)

Now, on comparing equation (2) and equation (3):

(2p)!p!p!×3p=(2p)!p!p!×3p$\\\\\Rightarrow \frac{(2p)!}{p!\;p!}\times 3^{p}=\frac{(2p)!}{p!\;p!}\times 3^{p}\\\\$

$\\\Rightarrow$  (2p)Cp × 312$\frac{1}{2}$ (2p – 1)Cp × 3p

Therefore, the coefficient of ap in the expansion of (1 + 3a)2p – 1 is half the coefficient of ap in the expansion of (1 + 3a)2p

Q.12: For what values of ‘m’ the coefficient of x2 in the expansion of (1+ 2x)k is 140.

Sol.

Since, The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × (b)r

Now, on substituting n = m, a =1 and b = 2x we will get:

Tr + 1 = mCr × (1)m – r × (2x)r

$\Rightarrow$ Tr + 1 = mCr × (2)r × (x)r . . . . . . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x2, we will get r = 2

Therefore, from equation (1):

T 2 + 1 = mC2 × (2)2 × (x)2

Since, the coefficient of x2 = 140   [GIVEN]

Therefore,    140 = mC2 × (2)2

84=[m!2!(m2)!]×4$\Rightarrow 84=\left [ \frac{m!}{2!\;\;(m-2)!} \right ]\times 4\\\\$ 21=m×(m1)(m2)![2×1]×(m2)!$\Rightarrow 21=\frac{m\times (m-1)\;\;(m-2)!}{[2\times 1]\times (m-2)!}\\\\$ 42=m2m$\Rightarrow 42 = m^{2}-m$

Therefore, m2 – m – 42 = 0

Now, by splitting of middle term method, the roots of this quadratic equation are:

$\Rightarrow$m2 – (7 – 6)m – 42 = 0

$\Rightarrow$m2 – 7m + 6m – 42 = 0

$\Rightarrow$ m(m – 7) +6(m – 7) = 0

i.e.    (m + 6) (m – 7) = 0

Therefore, m = 7 or m = -6

Hence, for the coefficient of x2 in the expansion of (1+ 2x)k to be 140, the value of ‘m’ should be 7.

Q.13: In the expansion of (2x+3y)9, Find the middle terms.

Sol.

Here, n = 9

When n is odd then the middle terms in the expansion of (a + b)n are given by:

middle terms = (n+12)thtermand(n+12+1)thterm$\left ( \frac{n+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{n+1}{2} +1 \right )^{th}term$

Therefore, the middle terms in the expansion of (2x+3y)9 are:

(9+12)thtermand(9+12+1)thterm$\Rightarrow \left ( \frac{9+1}{2}\right )^{th}term\;\;and\;\;\left ( \frac{9+1}{2} +1 \right )^{th}term$

$\Rightarrow$ 5th term and 6th term

Now, 5th and 6th terms in the expansion of (2x+3y)9 are:

T5 = T4 + 1 = 9C4 × (2x)9 – 4 × (3y)4

$\Rightarrow$   T5 = 9C4 × (2x)5 × (3y)4

T5=9!4!5!×(2x)5×(3y)4$\Rightarrow T_{5}=\frac{9!}{4!\; 5!}\times (2x)^{5}\times (3y)^{4}$ T5=9×8×7×6×5!4×3×2×1×5!×32x5×81y4$\Rightarrow T_{5}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}\times 32x^{5}\times 81y^{4}\\\\$

T5=126×2592x5y4=414720x5y4$\Rightarrow T_{5} = 126\times 2592\;x^{5}y^{4}\boldsymbol{=414720\;x^{5}y^{4}}\\$

Therefore, 5th term = 414720 x5 y4

Now, T6 = T5 + 1 = 9C5 × (2x)9 – 5 × (3y)5

$\Rightarrow$ T6 = 9C5 × (2x)4 × (3y)5

T6=9!5!×4!×(2x)4×(3y)5$\Rightarrow T_{6}=\frac{9!}{5!\times 4!}\times (2x)^{4}\times (3y)^{5}$ T6=9×8×7×6×5!5!×(4×3×2×1)×16x4×243y5$\Rightarrow T_{6}=\frac{9\times 8\times 7\times 6\times 5!}{5!\times (4\times 3\times 2\times 1)}\times 16x^{4}\times 243y^{5}\\\\$ T6=126×3888x4y5=489888x4y5$\Rightarrow T_{6} = 126\times 3888\;x^{4}y^{5}\boldsymbol{=489888\;x^{4}y^{5}}\\$

Therefore, 6th term = 489888 x4 y5

Hence, 5th and 6th terms are the middle terms in the expansion of (2x+3y)9

And also, 5th term = 414720 x5 y4 and 6th term = 489888 x4 y5

Q.14: Find the Coefficient of x6 in expansion of (x + 3)11

Sol.

Since, The general term in the expansion of (a + b)is given by:  Tr + 1 = nCr × (a)n – r × br

Therefore, on substituting n = 11, a = x and b = 3 in the above expression we will get:

Tr + 1 = 11Cr × (x)11 – r × 3r . . . . . . . (1)

Now, on comparing the coefficient of x in equation (1) with x7, we will get:

i.e.         (x)11 – r = x6

(or)         11 – r = 6

Therefore, r = 5

Now, on substituting r in equation (1) we will get:

T5 + 1 = 11C5 × (x)11 – 5 × 35

T6=11!5!×6!×x6×35$\Rightarrow T_{6}=\frac{11!}{5!\times 6!}\times x^{6}\times 3^{5}$ T6=11×10×9×8×7×6!5×4×3×2×1×6!×243x6$\Rightarrow T_{6}=\frac{11\times 10\times 9\times 8\times 7\times 6!}{5\times 4\times 3\times 2\times 1\times 6!}\times 243\;x^{6}\\\\$

T6=462×243x6=112266x6$\Rightarrow T_{6} = 462\times 243\;x^{6}\boldsymbol{=112266\;x^{6}}$

Therefore, the Coefficient of x6 in the expansion of (x + 3)11 = 112266

Q.15: Write the general term in the expansion of (x2y3 – x3 y2)7

Sol.

The general term in the expansion of (a + b)n is given by: Tr + 1 = nCr × (a)n – r × br

Now, on substituting n = 7, a = x2y3 and b = x3 y2 we will get:

Tr+1 = 7Cr × (x2y3)7 – r × (x3 y2)r

$\Rightarrow$ Tr + 1 = 7Cr × (x2)7 – r × (y3)7 – r × (x)3 r × (y)2 r

$\Rightarrow$ Tr + 1 = 7Cr × (x)14 – 2 r × (x)3 r× (y)2 r× (y)21 – 3r

$\Rightarrow$ Tr + 1 = 7Cr × (x)14 – 2 r + 3r  ×  (y)21 – 3r + 2 r

i.e.  Tr + 1 = 7Cr × (x)14 + r  ×  (y)21 – r

Therefore, the general term in the expansion of (x2y3 – x3 y2)7:

Tr + 1 = 7Cr × (x)14 + r × (y)21 – r