The sum of the first 20 terms of the series 1 | vedclass

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Question

(D) The given series is $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \dots$ The $n$-th term $T_n$ can be written as $T_n = \frac{2^n - 1}{2^{n-1}}

Vedclass · Accepted Answer

$38 + \frac{1}{2^{19}}$

Vedclass · Answer

(D) The given series is $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \dots$ The $n$-th term $T_n$ can be written as $T_n = \frac{2^n - 1}{2^{n-1}} = 2 - \frac{1}{2^{n-1}}$ for $n \ge 1$. The sum of the first $20$ terms is $S_{20} = \sum_{n=1}^{20} (2 - \frac{1}{2^{n-1}})$. $S_{20} = \sum_{n=1}^{20} 2 - \sum_{n=1}^{20} \frac{1}{2^{n-1}}$. $S_{20} = 2(20) - (1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{19}})$. The second part is a geometric progression with $a = 1$,$r = \frac{1}{2}$,and $n = 20$. Sum $= \frac{1(1 - (1/2)^{20})}{1 - 1/2} = 2(1 - \frac{1}{2^{20}}) = 2 - \frac{1}{2^{19}}$. Therefore,$S_{20} = 40 - (2 - \frac{1}{2^{19}}) = 38 + \frac{1}{2^{19}}$.

The sum of the first $20$ terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is?

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