Question

The sum of series $\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+.....$ up to $n$ terms is

• A. $n-\frac{4^n}{3}-\frac{1}{3}$
• B. $n+\frac{4^{-n}}{3}-\frac{1}{3}$
• C. $n+\frac{4^n}{3}-\frac{1}{3}$
• D. $n-\frac{4^{-n}}{3}-\frac{1}{3}$

Answer 1

Answer: (B)

$n+\frac{4^{-n}}{3}-\frac{1}{3}$

For $n=1$, we have
$n-\frac{4^n}{3}-\frac{1}{3}=1-\frac{4}{3}-\frac{1}{3}=-\frac{2}{3}$
$n+\frac{4^n}{3}-\frac{1}{3}=1+\frac{4}{3}-\frac{1}{3}=2$
$n-\frac{4^{-n}}{3}+\frac{1}{3}=1-\frac{4^{-1}}{3}+\frac{1}{3}=\frac{5}{4}$
Also, for $n=2$, we have
$n+\frac{4^{-n}}{3}-\frac{1}{3}=2+\frac{1}{48}-\frac{1}{3}=\frac{27}{16}$ and $\frac{3}{4}+\frac{15}{16}=\frac{27}{16}$
Hence, option (b) is correct.
ALTER We have,
$\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+.....$ to n terms
$=\frac{2^2-1}{2^2}+\frac{2^4-1}{2^4}+\frac{2^6-1}{2^6}+....$ to n terms.
$=(1-\frac{1}{2^2})+(1-\frac{1}{2^4})+(1-\frac{1}{2^6})+.....$ to n terms
$=n-\{\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+....\text{to n terms}\}$
$=n-\frac{1}{2^2}\left\{\frac{1-{\left(\frac{1}{2^2}\right)}^n}{1-\frac{1}{2^2}}\right\}$
$=n-\frac{1}{3}(1-4^{-n})$
$=n+\frac{4^{-n}}{3}-\frac{1}{3}$