Question

Let the coefficients of $x^n$ in ${(1+x)}^{2n}\&{(1+x)}^{2n-1}$ be P and Q, respectively, then ${\left(\frac{P+Q}{Q}\right)}^5=$

• A. 9
• B. 27
• C. 81
• D. none of these

Answer 1

The correct option is D none of these

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

The expansion${(1+x)}^{2n}$has$T_{r+1}={{}^{2n}C}_r{x}^r{(-1)}^{2n-r}={{}^{2n}C}_r{x}^r$

Co-efficient of$x^n={{}^{2n}C}_n$

We know that the co-efficient of$x^n$ in the expansion of${(1+x)}^{2n}$ is twice the coefficient of$x^n$ in the expansion${(1+x)}^{2n-1}$

$T_{r+1}={{}^{2n-1}C}_r{x}^r$

Co-efficient of$x^n={{}^{2n-1}C}_n$

We have${{}^{n}C}_r=\frac{n!}{r!(2n-1)!}$

Let $P={{}^{2n}C}_n=\frac{\left(2n\right)!}{n!(2n-n)!}=\frac{\left(2n\right)!}{n!(n!)}=\frac{2n(2n-1)!}{n!n(n-1)!}=\frac{2(2n-1)!}{n!(n-1)!}=2\alpha$

and$Q={{}^{2n-1}C}_n=\frac{(2n-1)!}{n!(n-1)!}=\alpha$

Therefore,$P=2Q$

Substituting,$P=2Q$ in ${\left(\frac{P+Q}{Q}\right)}^5={\left(\frac{2Q+Q}{Q}\right)}^5=3^5=243$