The sum of n terms of the following series 1 | vedclass

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(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k =

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$\frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$

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(C) The $k$-th term of the series is $T_k = 1 + x + x^2 + \dots + x^{k-1} = \frac{1 - x^k}{1 - x}$.The sum of $n$ terms is $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x}$.$S_n = \frac{1}{1 - x} \sum_{k=1}^{n} (1 - x^k) = \frac{1}{1 - x} \left[ \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right]$.$S_n = \frac{1}{1 - x} \left[ n - \frac{x(1 - x^n)}{1 - x} \right]$.$S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}$.

The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

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