Question

Find the coefficient of:
(i) x¹⁰ in the expansion of ${\left(2x^2-\frac{1}{x}\right)}^{20}$

(ii) x⁷ in the expansion of ${\left(x-\frac{1}{x^2}\right)}^{40}$

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

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

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

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

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

Answer 1

(i) Suppose x¹⁰ 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,
$$\begin{gathered} T_{r+1}={}^{20}C_r(2x^2{)}^{20-r}{\left(\frac{-1}{x}\right)}^r \\ =(-1{)}^r{}^{20}C_r\left(2^{20-r}\right)\left(x^{40-2r-r}\right) \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^{10},\mathrm{we}\mathrm{must}\mathrm{have}: \\ 40-3r=10 \\ \Rightarrow 3r=30 \\ \Rightarrow r=10 \\ \therefore \mathrm{Coefficient}\mathrm{of}{x}^{10}=(-1{)}^{10}{}^{20}C_{10}\left(2^{20-10}\right)={}^{20}C_{10}\left(2^{10}\right) \end{gathered}$$

(ii) Suppose x⁷ occurs at the (r + 1) th term in the given expression.

Then, we have:
$$\begin{gathered} T_{r+1}={}^{40}C_r{x}^{40-r}{\left(\frac{-1}{x^2}\right)}^r \\ =(-1{)}^r{}^{40}C_r{x}^{40-r-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^7,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 40-3r=7 \\ \Rightarrow 3r=40-7=33 \\ \Rightarrow r=11 \\ \therefore \mathrm{Coefficient}\mathrm{of}{x}^7=(-1{)}^{11}{}^{40}C_{11}=-{}^{40}C_{11} \end{gathered}$$

(iii) Suppose x⁻¹⁵ occurs at the (r + 1)th term in the given expression.
Then, we have:
$T_{r+1}={}^{10}C_r(3x^2{)}^{10-r}{\left(\frac{-a}{3x^3}\right)}^r$
$\Rightarrow T_{r+1}=(-1{)}^r{}^{10}C_r\left(3^{10-r-r}\right)\left(x^{20-2r-3r}\right)\left(a^r\right)$
$$\begin{gathered} \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^{-15},\mathrm{we}\mathrm{must}\mathrm{have}: \\ 20-5r=-15 \\ \Rightarrow 5r=20+15 \\ \Rightarrow r=7 \\ \therefore \mathrm{Coef}\mathrm{ficient}\mathrm{of}{x}^{-15}=(-1{)}^7{}^{10}C_7{3}^{10-14}\left(a^7\right)=-\frac{10\times 9\times 8}{3\times 2\times 9\times 9}{a}^7=-\frac{40}{27}{a}^7 \end{gathered}$$

(iv) Suppose x⁹ occurs at the (r + 1)th term in the above expression.

Then, we have:

$$\begin{gathered} T_{r+1}={}^{9}C_r(x^2{)}^{9-r}{\left(\frac{-1}{3x}\right)}^r \\ =(-1{)}^r{}^{9}C_r\left(x^{18-2r-r}\right)\left(\frac{1}{3^r}\right) \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^9,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 18-3r=9 \\ \Rightarrow 3r=9 \\ \Rightarrow r=3 \\ \therefore \mathrm{Coefficient}\mathrm{of}{x}^9=(-1{)}^3{}^{9}C_3\frac{1}{3^3}=-\frac{9\times 8\times 7}{2\times 9\times 9}=\frac{-28}{9} \end{gathered}$$

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

Then, we have:

$$\begin{gathered} T_{r+1}={}^{n}C_r{x}^{n-r}\frac{1}{x^r} \\ ={}^{n}C_r{x}^{n-2r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^m,\mathrm{we}\mathrm{must}\mathrm{have}: \\ n-2r=m \\ \Rightarrow r=(n-m)/2 \\ \therefore \mathrm{Coefficient}\mathrm{o}f{x}^m={}^{n}C_{(n-m)/2}=\frac{n!}{{\displaystyle \left(\frac{n-m}{2}\right)}!\left({\displaystyle \frac{n+m}{2}}\right)!} \end{gathered}$$

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

Then, we have:

$$\begin{gathered} (1-2x^3+3x^5){\left(1+\frac{1}{x}\right)}^8 \\ =\left(1-2x^3+3x^5\right)\left({}^{8}C_0+{}^{8}C_1\left(\frac{1}{x}\right)+{}^{8}C_2{\left(\frac{1}{x}\right)}^2+{}^{8}C_3{\left(\frac{1}{x}\right)}^3+{}^{8}C_4{\left(\frac{1}{x}\right)}^4+{}^{8}C_5{\left(\frac{1}{x}\right)}^5+{}^{8}C_6{\left(\frac{1}{x}\right)}^6+{}^{8}C_7{\left(\frac{1}{x}\right)}^7+{}^{8}C_8{\left(\frac{1}{x}\right)}^8\right) \\ x\mathrm{occurs}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expresssion}\mathrm{at}-2x^3.{}^{8}C_2\left(\frac{1}{x^2}\right)+3x^5.{}^{8}C_4{\left(\frac{1}{x}\right)}^4. \\ \therefore \mathrm{Coefficient}\mathrm{of}x=-2\left(\frac{8!}{2!6!}\right)+3\left(\frac{8!}{4!4!}\right)=-56+210=154 \end{gathered}$$

(vii)
Suppose a⁵ b⁷ occurs at the (r + 1)th term in the given expression.

Then, we have:

$$\begin{gathered} T_{r+1}={}^{12}C_r{a}^{12-r}(-2b{)}^r \\ =(-1{)}^r{}^{12}C_r\left(a^{12-r}\right)\left(b^r\right)\left(2^r\right) \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{a}^5{b}^7,\mathrm{we}\mathrm{must}\mathrm{have}: \\ 12-r=5 \\ \Rightarrow r=7 \\ \therefore \mathrm{Required}\mathrm{coefficient}=(-1{)}^7{}^{12}C_7\left(2^7\right)=-\frac{12\times 11\times 10\times 9\times 8\times 128}{5\times 4\times 3\times 2}=-101376 \end{gathered}$$