Question

Find the middle terms(s) in the expansion of:
(i) ${\left(x-\frac{1}{x}\right)}^{10}$

(ii) ${\left(1-2x+x^2\right)}^n$

(iii) ${\left(1+3x+3x^2+x^3\right)}^{2n}$

(iv) ${\left(2x-\frac{x^2}{4}\right)}^9$

(v) ${\left(x-\frac{1}{x}\right)}^{2n+1}$

(vi) ${\left(\frac{x}{3}+9y\right)}^{10}$

(vii) ${\left(3-\frac{x^3}{6}\right)}^7$

Answer 1

(i)
$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^{10} \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \therefore \mathrm{Middle}\mathrm{term}=\left(\frac{10}{2}+1\right)\mathrm{th}=6\mathrm{th}\mathrm{term} \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_6=T_{5+1} \\ ={}^{10}C_5{x}^{10-5}{\left(\frac{-1}{x}\right)}^5 \\ =-\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2} \\ =-252 \end{gathered}$$

(ii)
$$\begin{gathered} (1-2x+x^2{)}^n \\ =(1-x{)}^{2n} \\ n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \therefore \mathrm{Middle}\mathrm{term}=\left(\frac{2n}{2}+1\right)\mathrm{th}=(n+1)\mathrm{th}\mathrm{term} \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_{n+1}={}^{2n}C_n(-1{)}^n(x{)}^n \\ =\frac{\left(2n\right)!}{(n!{)}^2}(-1{)}^n{x}^n \end{gathered}$$

(iii)
$$\begin{gathered} (1+3x+3x^2+x^3{)}^{2n} \\ =(1+x{)}^{6n} \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \therefore \mathrm{Middle}\mathrm{term}=\left(\frac{6n}{2}+1\right)\mathrm{th}=(3n+1)\mathrm{th}\mathrm{term} \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_{3n+1} \\ ={}^{6n}C_{3n}{x}^{3n} \\ =\frac{\left(6n\right)!}{(3n!{)}^2}{x}^{3n} \end{gathered}$$

(iv)
$$\begin{gathered} {\left(2x-\frac{x^2}{4}\right)}^9 \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{middle}\mathrm{terms}\mathrm{are}\left(\frac{n+1}{2}\right)\mathrm{th}\mathrm{and}\left(\frac{n+1}{2}+1\right)\mathrm{th},\mathrm{i}.\mathrm{e}.5\mathrm{th}\mathrm{and}6\mathrm{th}\mathrm{terms}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_5=T_{4+1} \\ ={}^{9}C_4(2x{)}^{9-4}{\left(\frac{-x^2}{4}\right)}^4 \\ =\frac{9\times 8\times 7\times 6}{4\times 3\times 2}\times 2^5\frac{1}{4^4}{x}^{5+8} \\ =\frac{63}{4}{x}^{13} \\ \mathrm{And}, \\ {T}_6=T_{5+1} \\ ={}^{9}C_5(2x{)}^{9-5}{\left(\frac{-x^2}{4}\right)}^5 \\ =-\frac{9\times 8\times 7\times 6}{4\times 3\times 2}\times 2^4\frac{1}{4^5}{x}^{4+10} \\ =-\frac{63}{32}{x}^{14} \end{gathered}$$

(v)
$$\begin{gathered} {\left(x-\frac{1}{x}\right)}^{2n+1} \\ \mathrm{Here},\left(2n+1\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{middle}\mathrm{terms}\mathrm{are}\left(\frac{2n+1+1}{2}\right)\mathrm{th}\mathrm{and}\left(\frac{2n+1+1}{2}+1\right)\mathrm{th}\mathrm{i}.\mathrm{e}.(n+1)\mathrm{th}\mathrm{and}(n+2)\mathrm{th}\mathrm{terms}. \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ {T}_{n+1} \\ ={}^{2n+1}C_n{x}^{2n+1-n}\times \frac{(-1{)}^n}{x^n} \\ =(-1{)}^n{}^{2n+1}C_{n}x \\ \mathrm{And}, \\ {T}_{n+2}=T_{n+1+1} \\ ={}^{2n+1}C_{n+1}{x}^{2n+1-n-1}\frac{(-1{)}^{n+1}}{x^{n+1}} \\ =(-1{)}^{n+1}{}^{2n+1}C_{n+1}\times \frac{1}{x} \end{gathered}$$

(vi)
$$\begin{gathered} {\left(\frac{x}{3}+9y\right)}^{10} \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{middle}\mathrm{term}\mathrm{is}\left(\frac{10}{2}+1\right)\mathrm{th},\mathrm{i}.\mathrm{e}.,6\mathrm{th}\mathrm{term}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_6=T_{5+1} \\ ={}^{10}C_5{\left(\frac{x}{3}\right)}^{10-5}(9y{)}^5 \\ =\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times \frac{1}{3^5}\times 9^5\times x^5{y}^5 \\ =61236x^5{y}^5 \end{gathered}$$

(vii)
$$\begin{gathered} {\left(3-\frac{x^3}{6}\right)}^7 \\ \mathrm{Here},n\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{middle}\mathrm{terms}\mathrm{are}\left(\frac{7+1}{2}\right)\mathrm{th}\mathrm{and}\left(\frac{7+1}{2}+1\right)\mathrm{th},\mathrm{i}.\mathrm{e}.,4\mathrm{th}\mathrm{and}5\mathrm{th}\mathrm{terms}. \\ \mathrm{Now},\mathrm{we}\mathrm{have} \\ {T}_4=T_{3+1} \\ ={}^{7}C_3{\left(3\right)}^{7-3}{\left(\frac{-x^3}{6}\right)}^3 \\ =-\frac{105}{8}{x}^9 \\ \mathrm{And}, \\ {T}_5=T_{4+1} \\ ={}^{9}C_4{\left(3\right)}^{9-4}{\left(\frac{-x^3}{6}\right)}^4 \\ =\frac{7\times 6\times 5}{3\times 2}\times 3^5\times \frac{1}{6^4}{x}^{12} \\ =\frac{35}{48}{x}^{12} \end{gathered}$$