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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2015 · Mathematics · Sequences and Series

The value of (3)/(4)+(15)/(16)+(63)/(64)+ upto n terms is

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

Options

  1. An−4n3−13n-\frac{4^{n}}{3}-\frac{1}{3}
  2. Bn+4−n3−13\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}
  3. Cn+4n3−13n+\frac{4^{n}}{3}-\frac{1}{3}
  4. Dn−4−n3+13\mathrm{n}-\frac{4^{-\mathrm{n}}}{3}+\frac{1}{3}

Correct answer

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

Step-by-step solution

Consider 34+1516+6364+…\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\ldots upto n\mathrm{n} 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 n term =\left(1-\frac{1}{2^{2}}\right)+(1−124)+(1−126)+…\left(1-\frac{1}{2^{4}}\right)+\left(1-\frac{1}{2^{6}}\right)+\ldots upton terms =(1+1+1+….=(1+1+1+\ldots. upto n\mathrm{n} terms )−(122+124+126+…)-\left(\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+\ldots\right. upto n terms )=n−122[1−(122)n1122]=n−13(1−4−n)=n+4−n3−13)=\mathrm{n}-\frac{1}{2^{2}}\left[\frac{1-\left(\frac{1}{2^{2}}\right)^{\mathrm{n}}}{1 \frac{1}{2^{2}}}\right]=\mathrm{n}-\frac{1}{3}\left(1-4^{-\mathrm{n}}\right)=\mathrm{n}+\frac{4^{-\mathrm{n}}}{3}-\frac{1}{3}
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