Question

Value of ${\sum }_{k=1}^n\left(C_k\right)\left(C_{k-1}\right)$ is

• A. ${{}^{2n}C}_n$
• B. $\frac{1}{2}\left({{}^{2n+2}C}_{n+1}\right)-{{}^{2n}C}_n$
• C. ${{}^{2n}C}_{n+2}$
• D. none of these

Answer 1

Answer: (B)

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

The following series reduces to
${}^n{C}_0{}^n{C}_1+{}^n{C}_1{}^n{C}_2+{}^n{C}_2{}^n{C}_3+....{}^{2n}{C}_{n-1}.{}^{2n}{C}_n$
Now 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, $r=n-1$ and $m=n$
Hence, ${}^{2n}{C}_{n-1}$
Now $\frac{1}{2}\left({}^{2n+2}{C}_{n+1}\right)-{}^{2n}{C}_n$
$=\frac{1}{2}[\frac{(2n+2)(2n+1)}{(n+1{)}^2}.{}^{2n}{C}_n-{}^{2n}{C}_n]$
$={}^{2n}{C}_n[\frac{2n+1}{n+1}-1]$
$={}^{2n}{C}_n.\frac{n}{n+1}$
$={}^{2n}{C}_{n-1}$