Question

$\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.....$ upto n terms equal to

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

Answer 1

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

Required sum
$=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.....+\frac{2^n-1}{2^n}$
$=(1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})+.....+(1-\frac{1}{2^n})$
$=(1+1+1+.........\text{n times})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+........+\frac{1}{2^n})$
$=n-\frac{1/2[1-(1/2{)}^n]}{1-1/2}=n-1+\frac{1}{2^n}$
Hence option 'C' is correct choice.