Question

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

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

Answer 1

Answer: (B)

$S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\dots ..n$ terms
$=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+\dots \dots \dots ..+\left(1-\frac{1}{2^n}\right)=n-\left(\frac{1}{2}+\frac{1}{2^2}+\dots \dots \cdot \frac{1}{2^n}\right)$
$=n-\frac{1}{2}\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\left(\frac{1}{2}\right)}=n+2^{-n}-1$