The sum of the first n terms of the series fr | vedclass

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Question

(c) The sum of the first $n$ terms is

${S_n} = \left( {1 - \frac{1}{2}} \right) + \left( {1 - \frac{1}{{{2^2}}}} \right) + \left( {1 - \frac{1}{{{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) The sum of the first $n$ terms is

${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 ${S_2} = 2 + {2^{ - 2}} - 1 = \frac{5}{4}$.

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

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