Sequences and Series 4 Question 9
9.
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
(1988, 2M)
(a) $2^{n}-n-1$
(b) $1-2^{-n}$
(c) $n+2^{-n}-1$
(d) $2^{n}+1$
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Answer:
Correct Answer: 9. (c)
Solution:
- Sum of the $n$ terms of the series $\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\ldots$ upto $n$ terms can be written as
$ \begin{aligned} & (1-\frac{1}{2})+(1-\frac{1}{4})+(1-\frac{1}{8})+(1-\frac{1}{16})+\ldots \text { upto } n \text { terms } \\ &=n-(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+n \text { terms } )\\ &=n-\frac{\frac{1}{2}( 1-\frac{1}{2^{n}})}{1-\frac{1}{2}}=n+2^{-n}-1 \end{aligned} $
