Question

The sum to $50$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+....$ is equal to

• A. $2^{50}-51$
• B. $1-2^{-50}$
• C. $2^{-50}+49$
• D. $2^{50}-1$

Answer 1

The correct option is C $2^{-50}+49$

$T_n=\frac{2^n-1}{2^n}=1-{\left(\frac{1}{2}\right)}^n$
$\therefore S_n=n-\sum _{n=1}^n{\left(\frac{1}{2}\right)}^n=n-\frac{\frac{1}{2}.[1-(1/2{)}^n]}{1-1/2}$
$=n+2^{-n}-1$.