Question

If $S_n={\sum }_{r=1}^n\frac{1+2+2^2+..Sumtorterms}{2^r}$, then $S_n$ is equal to

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

Answer 1

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

$\to S_n=\sum _{r=1}^n\left(\frac{\frac{2^r-1}{2-1}}{2^r}\right)$ sum of G.P

$\Rightarrow S_n=\sum _{r=1}^n(1-2^{-r})$

$\Rightarrow S_n=n-\sum _{r=1}^n{2}^{-r}$

$\Rightarrow S_n=n-\left(2^{-1}\right(\frac{2^{-n}-1}{2^{-1}-1}\left)\right)$

$\Rightarrow S_n=n-\left(\frac{1}{2}\right(\frac{1-2^n}{2^n\left(\frac{-1}{2}\right)}\left)\right)$

$\Rightarrow S_n=n+\left(\frac{1-2^n}{2^n}\right)=n-1+\frac{1}{2^n}$

$\Rightarrow \left(C\right)$