Sum of first n terms of series frac{1}{2} + frac{3}{4} + frac{7}{8} + frac{{15}}{{16}} + ......... i

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8. Sequences and Series

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

${2^n} - n - 1$

$1 - {2^{ - n}}$

$n + {2^{ - n}} - 1$

${2^n} - 1$

(IIT-1988)

Solution

${S_n} = \left( {1 – \frac{1}{2}} \right) + \left( {1 – \frac{1}{{{2^2}}}} \right) + \left( {1 – \frac{1}{{{2^3}}}} \right) + \left( {1 – \frac{1}{{{2^4}}}} \right)+$$ …… + \left( {1 – \frac{1}{{{2^n}}}} \right)$

$ = n – \left\{ {\frac{1}{2} + \frac{1}{{{2^2}}} + ….. + \frac{1}{{{2^n}}}} \right\}$

= $n – \frac{1}{2}\left( {\frac{{1 – \frac{1}{{{2^n}}}}}{{1 – \frac{1}{2}}}} \right) = n – \left( {1 – \frac{1}{{{2^n}}}} \right) = n – 1 + {2^{ – n}}$.

Trick : Check for $n = 1,\;2\;$

$i.e.$ ${S_1} = \frac{1}{2},\;{S_2} = \frac{5}{4}$

and $(c)$ $ \Rightarrow {S_1} = \frac{1}{2}$

and ${S_2} = 2 + {2^{ – 2}} – 1 = \frac{5}{4}$.

Standard 11

Mathematics