Question

If $s_n={\sum }_{n=1}^n\frac{1+2+2^2+...tonterms}{2^n}$ 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 $n-1+\frac{1}{2^n}$

Given that $s_n={\sum }_{n=1}^n\frac{1+2+2^2+...2^{n-1}}{2^n}$
$\Rightarrow s_n={\sum }_{n=1}^n\frac{1}{2^n}\left(\frac{2^n-1}{2-1}\right)={\sum }_{n=1}^{n}1-{\sum }_{n=1}^n\left(\frac{1}{2^n}\right)$ ...[ sum of G.P. series ]
$\Rightarrow s_n=n-\frac{1}{2}\left(\frac{1-{\left(1/2\right)}^n}{1-\left(1/2\right)}\right)=n-1+\frac{1}{2^n}$ ...[ sum of G.P. series ]
Ans: C