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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Sequences and Series

NDA 2017 · Mathematics · Sequences and Series

The sum of the first n terms of the series...

The sum of the first n terms of the series \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$. is equal to

Options

  1. A2n−n−12^{\mathrm{n}}-\mathrm{n}-1
  2. B1−2−n1-2^{-n}
  3. C1−2−n1-2^{-n}
  4. D2n−12^{\mathrm{n}}-1

Correct answer

C. 1−2−n1-2^{-n}

Step-by-step solution

12+34+78+1516+…..=(1−12)+(1−14)+(1−18)+(1−116)+……=(1+1+1+….n)−(12+14+18+116+…..)=n−(12+122+123+124+…..)=n−12(1−12n)1−12=n−12(1−12n)12=n−1+2−n=2−n+n−1\begin{aligned} & \frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots.. \\ & =\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+\left(1-\frac{1}{16}\right)+\ldots \ldots \\ & =(1+1+1+\ldots. \mathrm{n})-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots..\right) \\ & =\mathrm{n}-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\ldots..\right) \\ & =\mathrm{n}-\frac{\frac{1}{2}\left(1-\frac{1}{2^{\mathrm{n}}}\right)}{1-\frac{1}{2}} \\ & =\mathrm{n}-\frac{\frac{1}{2}\left(1-\frac{1}{2^{\mathrm{n}}}\right)}{\frac{1}{2}} \\ & =\mathrm{n}-1+2^{-\mathrm{n}} \\ & =2^{-\mathrm{n}}+\mathrm{n}-1\end{aligned}
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