In the expansion of the following expression $1 + (1 + x) + (1 + x)^{2} + . . . . + (1 + x)^{n}$ , the coefficient of $x^{k} (0 \leq k \leq n)$ is

^{n + 1} C_{k + 1}

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^{n} C_{k}

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^{n} C_{n - k - 1}

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none of these

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Solution

The correct option is A $^{n + 1} C_{k + 1}$
Given expression $1 + (1 + x) + (1 + x)^{2} + . . . . + (1 + x)^{n}$
The expression is in G. P. and total number of terms are $n + 1$
$E = 1 + (1 + x) + (1 + x)^{2} + . . . . + (1 + x)^{n}$
$E = \frac{(1 + x)^{n + 1} - 1}{(1 + x) - 1}$
$E = x^{- 1} {(1 + x)^{n + 1} - 1}$
$∴$ The coefficient of $x^{k}$ in E $=$ The coefficient of $x^{k + 1}$ in $x^{- 1} {(1 + x)^{n + 1} - 1} =^{n + 1} C_{k + 1}$

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