Question

If $S_n=\frac{1}{n+1}{\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)$, then $2S_{n+1}-S_n$ equals

• A. $=2^n\left[\frac{3n-2}{(n+1)(n+2)}\right]$
• B. $=2^n\left[\frac{3n+2}{(n+1)(n+2)}\right]$
• C. $=2^n\left[\frac{3n+2}{(n-1)(n+2)}\right]$
• D. 0

Answer 1

The correct option is B $=2^n\left[\frac{3n+2}{(n+1)(n+2)}\right]$

$\sum {}^n{C}_i=2^n$ ....sum of binomial coefficients of $(1+x{)}^n$
Hence

$S_n=\frac{2^n}{n+1}$

Now

$S_{n+1}=\frac{2^{n+1}}{n+2}$

$2S_{n+1}-S_n$

$=\frac{2^{n+2}}{n+2}-\frac{2^n}{n+1}$
$=2^n[\frac{4}{n+2}-\frac{1}{n+1}]$
$=2^n\left[\frac{4n+4-n-2}{(n+1)(n+2)}\right]$
$=2^n\left[\frac{3n+2}{(n+1)(n+2)}\right]$