Question

The sum of the series $1+{\frac{1}{2}}^n{C}_1+{\frac{1}{3}}^n{C}_3+.........+{\frac{1}{n+1}}^n{C}_n$ is equal to

• A. $\frac{2^{n+1}-1}{n+1}$
• B. $\frac{3(2^n-1)}{2n}$
• C. $\frac{2^n+1}{n+1}$
• D. $\frac{2^n+1}{2n}$

Answer 1

The correct option is A $\frac{2^{n+1}-1}{n+1}$

$1+{\frac{1}{2}}^n{C}_1+{\frac{1}{3}}^n{C}_2+......+{\frac{1}{n+1}}^n{C}_n$

$=\frac{1}{n+1}\left[\right(n+1)+\frac{(n+1)n}{2!}+\frac{(n+1)n(n-1)}{3!}++1]$

$=\frac{1}{n+1}{[}^{n+1}{C}_1{+}^{n+1}{C}_2+{+}^{n+1}{C}_{n+1}]$

$=\frac{1}{n+1}{[}^{n+1}{C}_0{+}^{n+1}{C}_1{+}^{n+1}{C}_2+{+}^{n+1}{C}_{n+1}-1]$ $[\because {}^{n+1}{C}_0=1]$

$=\frac{1}{n+1}(2^{n+1}-1)$