Question

If ${(1+x)}^n=\sum _{r=0}^n{C}_r{x}^r$ the the value of $C_0^2+C_1^2+C_2^2+...C_n^2$ equals

• A. $\frac{2n!}{n!n-1!}$
• B. ${\frac{2n!}{(n-1!)}}^2$
• C. $\frac{\mathrm{1.3.5...}(2n-1)2^n}{n!}$
• D. None of these

Answer 1

Answer: (C)

$\frac{\mathrm{1.3.5...}(2n-1)2^n}{n!}$

Consider the following series
.${}^m{C}_0{}^n{C}_r+{}^m{C}_1{}^n{C}_{r-1}+...{}^m{C}_r{}^n{C}_0$
=Coefficient of $x^r$ in $(1+x{)}^m.(x+1{)}^n$
=coefficient of $x^r$ in $(1+x{)}^{m+n}$
$={}^{m+n}{C}_r$
In the above case, $r=n$ and $m=n$
Hence
${}^{2n}{C}_n$
$=\frac{2n!}{n{!}^2}$
$=\frac{\mathrm{1.3.5..}(2n-1)}{n!}{.2}^n$