Question

Find the coefficient of:

$\left(i\right)x^{10}$ in the expansion of ${(2x^2-\frac{1}{x})}^{20}$

$\left(ii\right)x^7$ in the expansion of ${(x-\frac{1}{x^2})}^{40}$

$\left(iii\right)x^{-15}$ in the expansion of ${(3x^2-\frac{a}{3x^3})}^{10}$

$\left(iv\right)x^9$ in the expansion of ${(x^2-\frac{1}{3x})}^9$

$\left(v\right)x^m$ in the expansion of ${(x+\frac{1}{x})}^n$

(vi)x in the expansion of $(1-2x^3+3x^5){(1+\frac{1}{x})}^8$

$\left(vii\right)a^5{b}^7$ in the expansion of $(a-2b{)}^{12}.$

(viii)x in the expansion of $(1-3x+7x^2)(1-x{)}^{16}$

Answer 1

(i)Suppose $x^{10}$ occurs in the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^n{C}_r{x}^{n-r}{a}^r$

Here,

$T_{r+2}={}^{20}{C}_r(2x^2{)}^{20-r}{\left(\frac{-1}{x}\right)}^r$

$(-1{)}^r{}^{20}{C}_r(2^{20-r}\left)\right(x^{40-2r-r})$

For this term to contain $x^{10}$ we must have:

40-3r=10

$\Rightarrow 3r=30$

$\Rightarrow r=10$

$\therefore$ coefficient of $x^{10}=(-1{)}^{10}{}^{20}{C}_{10}(2^{20-10})$

$={}^{20}{C}_{10}\left(2^{10}\right)$

(ii)Suppose $x^7$ occurs at the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^{40}{C}_r{x}^{40-r}{\left(\frac{-1}{x^2}\right)}^r$

$=(-1{)}^r{}^{40}{C}_r{x}^{40-r-2r}$

For this term to contain $x^7$, we must have:

=40-3r=7

$\Rightarrow 3r=40-7=33$

$\Rightarrow r=11$

$\therefore$ Coefficient of $x^7=(-1{)}^{11}{}^{40}{C}_{11}=-{}^{40}{C}_{11}$

(iii)Suppose $x^{-15}$ occurs in the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^{10}{C}_r(3x^2{)}^{10-r}{\left(\frac{-a}{3x^3}\right)}^r$

$\Rightarrow T_{r+1}=(-1{)}^r{}^{10}{C}_r(3^{10-r-r}\left)\right(x^{20-2r-3r}\left)\right(d^r)$

For this term to contain $x^{-15}$, we must have:

=20-5r =-15

$\Rightarrow 5r=20+15$

$\Rightarrow r=7$

$\therefore$ Coefficient of $x^{-15}=(-1{)}^7{}^{10}{C}_7{3}^{10-14}$

$\left(a^7\right)=-\frac{10\times 9\times 8}{3\times 2\times 9\times 9}{a}^7=-\frac{40}{27}{a}^7$

(iv)Suppose $x^9$ occurs at the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^9{C}_r(x^2{)}^{9-r}{\left(\frac{-1}{3x}\right)}^r=(-1{)}^r{}^9{C}_r\left(x^{18-2r-4}\right)\left(\frac{1}{3^r}\right)$

For this term to contain $x^9$, we must have:

18-3r =9

$\Rightarrow 3r=9$

$\Rightarrow r=3$

$\therefore$ Coefficient of $x^9=(-1{)}^3{}^9{C}_3\frac{1}{3^3}$

$=-\frac{9\times 8\times 7}{2\times 9\times 9}=\frac{-28}{9}$

(v) Suppose $x^m$ occurs at the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^n{C}_r{x}^{n-r}\frac{1}{x^r}$

${=}^{n{C}_r}{x}^{n-2r}$

For this term to contain $x^m$, we must have:

n-2r =m

$\Rightarrow r=\frac{(n-m)}{2}$

$\therefore Coefficientof{x}^m=\frac{{}^n{C}_{(n-m)}}{2}=\frac{n!}{\left(\frac{n-m}{2}\right)!\left(\frac{n+m}{2}\right)!}$

(vi)Suppose x occurs at the (r+1)th term in the given expression.

Then, we have:

$(1-2x^3+3x^5){(1+\frac{1}{x})}^8(1-2x^3+3x^5)\left({}^8{C}_0{+}^8{C}_1\right(\frac{1}{x}\left){+}^8{C}_2\right(\frac{1}{x}{)}^2{+}^8{C}_3\left(\frac{1}{x}{)}^3{+}^8{C}_4\right(\frac{1}{x}{)}^4{+}^8{C}_5\left(\frac{1}{x}{)}^5{+}^8{C}_6\right(\frac{1}{x}{)}^6{+}^8{C}_7(\frac{1}{x}{)}^7{}^{}{+}^8{C}_8\left(\frac{1}{x}{)}^8\right)$

x occurs in the above expression at $-2x^3\times {}^8{C}_2\left(\frac{1}{x^2}\right)+3x^5\times {}^8{C}_4{\left(\frac{1}{x}\right)}^4$

$\therefore$ Coefficient of $x=-2\left(\frac{8!}{2!6!}\right)+3\left(\frac{8!}{4!4!}\right)=56+216=154$

$\left(vii\right)a^5{b}^7$ in the expansion of $(a-2b{)}^{12}$

Suppose $a^5{b}^7$ occurs at the (r+1)th term in the given expression.

Then, we have:

$T_{r+1}={}^{12}{C}_r{a}^{12-r}(-2b{)}^r=(-1\left)r{}^{12}{C}_r\right(a^{12-r}\left)\right(b^r\left)\right(2^r)$

For this term to contain $a^5{b}^7$, we must have:

$12-r=5,\Rightarrow r=7$

$\therefore$ Required coefficient $=(-1{)}^7{}^{12}{C}_7(2^7)=\frac{-12\times 11\times 10\times 9\times 8\times 128}{5\times 4\times 3\times 2}=-101376$

(viii)Suppose x occurs at the (r+1)th term in the given expression.

Then, we have:

$(1-3x+7x^2)(1-x{)}^{16}=(1-3x+7x^2\left){(}^{16}{C}_0{+}^{16}{C}_1\right(-x\left){+}^{16}{C}_2\right(-x{)}^2{+}^{16}{C}_3(-x{)}^3{+}^{16}{C}_4(-x{)}^4{}^{}{+}^{16}{C}_5(-x{)}^5{+}^{16}{C}_6(-x{)}^6{+}^{16}{C}_7(-x{)}^7{+}^{16}{C}_8(-x{)}^8{+}^{16}{C}_9(-x{)}^9{+}^{16}{C}_{10}(-x{)}^{10}{+}^{16}{C}_{11}(-x{)}^{11}{}^{}{+}^{16}{C}_{12}(-x{)}^{12}{+}^{16}{C}_{13}(-x{)}^{13}{+}^{16}{C}_{14}(-x{)}^{14}{+}^{16}{C}_{15}(-x{)}^{15}{+}^{16}{C}_{16}(-x{)}^{16})$

x occurs in the above expression at ${}^{16}{C}_1(-x)-3x{}^{16}{C}_0$

$\therefore$ Coefficient of $x=\left(\frac{16!}{1!15!}\right)-3\left(\frac{16!}{0!16!}\right)=-16-3=-19$