Question

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

• A. 243
• B. 27
• C. 81
• D. none of these

Answer 1

The correct option is A 243

Co-efficient of $x^n$ is the expansion of $(1+x{)}^{2n}={{}^{2n}C}_n$
Coefficient of $x^n$ in the expansion of $(1+x{)}^{2n-1}={{}^{2n-1}C}_n$

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

As we know ${{}^{2n}C}_n=P=2{{}^{2n-1}C}_n=Q=1$

So $(\frac{P+Q}{Q}{)}^5=(\frac{2+1}{1}{)}^5=(3{)}^5=243\therefore (\frac{P+Q}{Q}{)}^5=243$