Question

Find the middle term(s)in the expansion of:

$\left(i\right){(x-\frac{1}{x})}^{10}$

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

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

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

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

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

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

$\left(viii\right){(2ax-\frac{b}{x^2})}^{12}$

$\left(ix\right){(\frac{p}{x}+\frac{x}{p})}^9$

$\left(x\right){(\frac{x}{a}-\frac{a}{x})}^{10}$

Answer 1

$\left(i\right){(x-\frac{1}{x})}^{10}$

Here, n=10, which is even, $\therefore$ it has 11 terms

$\therefore$ middle term is $(\frac{n}{2}+1)=6^{th}$ term

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_6=T_{5+1}=(-1{)}^5{}^{10}{C}_5{x}^{10-5}{\left(\frac{1}{x}\right)}^5$

$=\frac{-10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times \frac{x^5}{x^5}$

$=-3\times 2\times 7\times 6$

=-252

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

Here, n is odd, $\therefore (1-2x+x^2)$ has n+1 =even term

$\therefore$ middle term is ${\left(\frac{n+1}{2}\right)}^{th}$ term

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

$T_{\frac{n+1}{2}}=T_{\frac{n}{2}}={}^n{C}_{\frac{n}{2}}(1-2x{)}^{n-\frac{n}{2}}(x^2{)}^{\frac{n}{2}}$

$=\frac{n!}{\frac{n}{2}!\frac{n}{2}!}(1-2x{)}^{\frac{n}{2}}{x}^{\frac{2n}{2}}$

$=\frac{\left(2n\right)!}{(n!{)}^2}(-1{)}^n{x}^n[\therefore (1-x{)}^n=1-nx]$

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

This expansion is $\left(\right(1+x{)}^3{)}^{2n}=(1+x{)}^{6n}$ Since 6n is even $\therefore$ it has 6n+1 =odd terms has middle term is

${(\frac{6n}{2}+1)}^{th}=(4n{)}^{th}$ term

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

$T_{4n}=T_{3n+1}={}^{6n}{C}_{3n}\left(1{)}^{6n-3n}\right(x{)}^{3n}$

$=\frac{\left(6n\right)!}{\left(3n\right)!\left(3n\right)!}{x}^{3n}[\because 1^{6n-3n}=1]$

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

Here, n is an odd number.

Therefore, the middle terms are $\left(\frac{n+1}{2}\right)thand(\frac{n+1}{2}+1)th$, i.e. 5th and 6th terms

Now, we have

$T_5=T_{4+1}$

$={}^9{C}_4(2x{)}^{9-4}{\left(\frac{-x^2}{4}\right)}^4$

$=\frac{9\times 8\times 7\times 6}{4\times 3\times 2}\times 2^5\frac{1}{4}{x}^{5+8}$

$=\frac{63}{4}{x}^{13}$

And,

$T_6=T_{5+1}$

$={}^9{C}_5(2x{)}^{9-5}{\left(\frac{-x^2}{4}\right)}^5$

$=-\frac{9\times 8\times 7\times 6}{4\times 3\times 2}\times 2^4\frac{1}{4^5}{x}^{4+10}$

$=\frac{63}{32}{x}^{14}$

$\left(v\right){(x-\frac{1}{x})}^{2n+1}$ 2n+1 is odd hence this expansion will have 2n+2 =even terms.

Hence, middle terms is $\frac{2n+1}{2}$

=n+1, n+2

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_{n+1}=T_{n+1}=(-1{)}^n{}^{2n+1}{C}_n(x{)}^{2n+1-n}{\left(\frac{1}{x}\right)}^n$

$=(-1{)}^n{}^{2n+1}{C}_n{x}^{n+1-n}$

$=(-1{)}^n{}^{2n+1}{C}_{n}x$

$T_{n+2}=T_{n+1+1}$

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

$=(-1{)}^{n+1}{}^{2n+1}{C}_{n+1}{x}^{-1}$

$=(-1{)}^{n+1}{}^{2n+1}{C}_{n+1}\frac{1}{x}$

$=(-1{)}^{n+1}{}^{2n+1}{C}_n\frac{1}{x}[\because {}^n{C}_r={}^n{C}_{r-1}]$

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

Here n=10, which is even, therefore it has 11 terms

$\therefore$ middle term is $(\frac{n}{2}+1)=6^4$ term

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_6=T_{5+1}=(-1{)}^5{}^{10}{C}_5{\left(\frac{x}{3}\right)}^{10-5}(9y{)}^5$

$=-\frac{10!}{5!5!}\times \frac{x^5}{3^5}\times 9^5\times y^5$

$=61236x^5{y}^5$

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

Here n=7, which is odd

$\therefore$ middle term is $\left(\frac{7+1}{2}\right)and(\frac{7+1}{2}+1)$

$=4^{th},5^{th}$ terms

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_4=T_{3+1}=(-1{)}^3{}^7{C}_3(3{)}^{7-2}{\left(\frac{x^3}{6}\right)}^3$

$=-\frac{7!}{3!4!}\times 3^4\times \frac{x^9}{6^2}$

$=-\frac{7\times 6\times 5}{3\times 2\times 1}\times 81\times \frac{x^9}{216}$

$=-\frac{105}{8}{x}^9$

And

$T_n=T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$T_3=T_{4+1}=(-1{)}^4{}^7{C}_4(3{)}^{7-4}{\left(\frac{x^3}{6}\right)}^4$

$=\frac{7!}{4!3!}\times 3^3\times \frac{x^{12}}{6^4}$

$=\frac{7\times 6\times 5}{3\times 2\times 1}\times 27\times \frac{x^{12}}{1296}$

$=\frac{35}{48}{x}^{12}$

$\left(viii\right){(2ax-\frac{b}{x^2})}^{12}$

Here, n is an even number.

$\therefore$ Middle term $=(\frac{12}{2}+1)th=7^{th}$ term

Now, we have

$T_7=T_{6+1}$

$={}^{12}{C}_6(2ax{)}^{12-5}{\left(\frac{-b}{x^2}\right)}^6$

$=\frac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1}\times {\left(\frac{2ab}{x}\right)}^6$

$=\frac{59136a^6{b}^6}{x^6}$

$\left(ix\right){(\frac{p}{x}+\frac{x}{p})}^9$

Here, n is an odd number

Therefore, the middle terms are $\left(\frac{9+1}{2}\right)thand(\frac{9+1}{2}+1)th,i.e.,5^{th}$ and $6^{th}$ terms

Now, we have

$T_5=T_{4+1}$

$={}^9{C}_4{\left(\frac{p}{x}\right)}^{9-4}{\left(\frac{x}{p}\right)}^4$

$=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}\times \left(\frac{p}{x}\right)$

$=\frac{126p}{x}$

And,

$T_6=T_{5+1}$

$={}^9{C}_5{\left(\frac{p}{x}\right)}^{9-5}{\left(\frac{x}{p}\right)}^5$

$=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}\times \left(\frac{x}{p}\right)$

$=\frac{126x}{p}$

$\left(x\right){(\frac{x}{a}-\frac{a}{x})}^{10}$

Here, n is an even number

$\therefore$ Middle term $=(\frac{10}{2}+1)th$ =6th term

Now, we have

$T_6=T_{5+1}$

$={}^{10}{C}_5{\left(\frac{x}{a}\right)}^{10-5}{\left(\frac{-a}{x}\right)}^5$

$=-\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 1}$

=-252