Question

The sum of the first n terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.....$ is
[IIT 1988; MP PET 1996; RPET 1996, 2000; Pb. CET 1994; DCE 1995, 96]

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

Answer 1

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

The sum of the first n terms is $S_n=(1-\frac{1}{2})+(1-\frac{1}{2^2})+(1-\frac{1}{2^3})+(1-\frac{1}{2^4})+.....+(1-\frac{1}{2^n})$
$=n-\{\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}\}$
$=n-\frac{1}{2}\left(\frac{1-\frac{1}{2^n}}{1-\frac{1}{2}}\right)=n-(1-\frac{1}{2^n})=n-1+2^{-n}.$
Trick : Check for n = 1, 2 i.e., $S_1=\frac{1}{2},S_2=\frac{5}{4}and\left(c\right)\Rightarrow S_1=\frac{1}{2}and{S}_2=2+2^{-2}-1=\frac{5}{4}.$