Question

If $C_r=\left(\begin{array}{c}n\\ r\end{array}\right)$, then sum of series $\frac{C_0^2}{1}+\frac{C_1^2}{2}+\frac{C_2^2}{3}+....$ upto $(n+1)$ terms is

• A. $\frac{1}{n+1}\left(\begin{array}{c}2n+1\\ n+1\end{array}\right)$
• B. $\frac{1}{2(n+1)}\left(\begin{array}{c}2n+1\\ n+1\end{array}\right)$
• C. $\frac{1}{n+1}\left(\begin{array}{c}2n+1\\ n\end{array}\right)$
• D. $\frac{(2n+1)!}{{(n+1)!}^2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{1}{n+1}\left(\begin{array}{c}2n+1\\ n+1\end{array}\right)$
C $\frac{1}{n+1}\left(\begin{array}{c}2n+1\\ n\end{array}\right)$
D $\frac{(2n+1)!}{{(n+1)!}^2}$

For $r\ge 0$
$\frac{1}{r+1}{C}_r^2=\frac{1}{r+1}\cdot \frac{n!}{r!(n-r)!}{C}_r$
$=\frac{1}{n+1}\frac{(n+1)!}{(r+1)!(n-r)!}{C}_r$
$=\frac{1}{n+1}\left(\begin{array}{c}n+1\\ n-r\end{array}\right)\left(\begin{array}{c}n\\ r\end{array}\right)$
Thus, $S=\frac{1}{n+1}{S}_1$ where
$S_1={\sum }_{r=0}^n\left(\begin{array}{c}n+1\\ n-r\end{array}\right)\left(\begin{array}{c}n\\ r\end{array}\right)$
$=$ number of ways of selecting $n$ persons out of $(n+1)$ men and $n$ women
$=$ number of ways of selecting $n$ persons out of $(2n+1)$ persons
$=\left(\begin{array}{c}2n+1\\ n+1\end{array}\right)=\frac{(2n+1)!}{n!(n+1)!}$
$\therefore$ $S=\frac{1}{n+1}\frac{(2n+1)!}{n!(n+1)!}=\frac{1}{(n+1)}\left(\begin{array}{c}2n+1\\ n+1\end{array}\right)=\frac{1}{n+1}\left(\begin{array}{c}2n+1\\ n\end{array}\right)=\frac{1}{n+1}\frac{(2n+1)!}{n!(n+1)!}=\frac{(2n+1)!}{(n+1){!}^2}$

Hence, options $A,C$ and $D$ are correct.