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Mathematics Question on Sequence and series

The value of 34+1516+6364+.....\frac{3}{4} + \frac{15}{16} +\frac{63}{64} +..... upto n terms is

A

n−4n3−13n - \frac{4^n}{3} - \frac{1}{3}

B

n+4−n3−13n + \frac{4^{-n}}{3} - \frac{1}{3}

C

n+4n3−13n + \frac{4^n}{3} - \frac{1}{3}

D

n−4−n3+13n - \frac{4^{-n}}{3} + \frac{1}{3}

Answer

n+4−n3−13n + \frac{4^{-n}}{3} - \frac{1}{3}

Explanation

Solution

Consider 34+1516+6364+...\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+... upto nn terms
=22−122+24−124+26−126=\frac{2^{2}-1}{2^{2}}+\frac{2^{4}-1}{2^{4}}+\frac{2^{6}-1}{2^{6}} upto nn terms
=(1−122)+(1−124)+(1−126)+...=\left(1-\frac{1}{2^{2}}\right)+\left(1-\frac{1}{2^{4}}\right)+\left(1-\frac{1}{2^{6}}\right)+... upto nn terms
=(1+1+1+… upto n terms )−(122+124+126+… upto n terms )=(1+1+1+\ldots \text { upto } n \text { terms })-\left(\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+\ldots \text { upto } n \text { terms }\right)
=n−122[1−(122)n1−122]= n -\frac{1}{2^{2}}\left[\frac{1-\left(\frac{1}{2^{2}}\right)^{ n }}{1-\frac{1}{2^{2}}}\right]
=n+4−n3−13= n +\frac{4^{- n }}{3}-\frac{1}{3}