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

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(C) The sum of the first $n$ terms is given by:$S_n = \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{2^2}\right) + \left(1 - \frac{1}{2^3}\right) +

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

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(C) The sum of the first $n$ terms is given by:$S_n = \left(1 - \frac{1}{2}\right) + \left(1 - \frac{1}{2^2}\right) + \left(1 - \frac{1}{2^3}\right) + \dots + \left(1 - \frac{1}{2^n}\right)$$S_n = \sum_{k=1}^{n} \left(1 - \frac{1}{2^k}\right) = n - \sum_{k=1}^{n} \left(\frac{1}{2}\right)^k$Using the sum formula for a geometric progression $\sum_{k=1}^{n} ar^{k-1} = a\frac{1-r^n}{1-r}$,we have:$S_n = n - \left[ \frac{1}{2} \left( \frac{1 - (1/2)^n}{1 - 1/2} \right) \right]$$S_n = n - \left( 1 - \frac{1}{2^n} \right) = n - 1 + 2^{-n}$Verification for $n=1$: $S_1 = 1 - 1 + 2^{-1} = 1/2$ (Correct).Verification for $n=2$: $S_2 = 2 - 1 + 2^{-2} = 1 + 1/4 = 5/4$ (Correct).

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

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