The sum $1 + (1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + . . . . n$ terms equals.

\frac{1 - x^{n}}{1 - x}

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

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

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None of these

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Solution

The correct option is C $\frac{n (1 - x) - x (1 - x^{n})}{(1 - x)^{2}}$
Let, $S_{n} = 1 + (1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + . . . . . . . . . . . . . . . . . . . . . . . . . . . n$ terms
So $S_{n - 1} = 1 + (1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + . . . . . . . . n - 1$ terms
$∴ T_{n} = S_{n} - S_{n - 1} = 1 + x + x^{2} + x^{3} + . . . . . . . . . . . . . . . . . . x^{n} = \frac{1 - x^{n}}{1 - x}$
Therefore, the required summation is,
$\sum T_{n} = \frac{1}{1 - x} (\sum 1 - \sum x^{n}) = \frac{1}{1 - x} (n - \frac{x (1 - x^{n})}{1 - x})$
$= \frac{n (1 - x) - x (1 - x^{n})}{(1 - x)^{2}}$
Hence, option 'C' is correct.

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