The sum of the series ${10.}^{n} C_{0} + 10^{2} .^{n} C_{1} + 10^{3} .^{n} C_{2} + . . {.10}^{n + 1} .^{n} C_{n}$ is

11^{n}

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{10.11}^{n}

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11^{n + 1}

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11^{n} - 1

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Solution

The correct option is D ${10.11}^{n}$
$\to 10 (^{n} C_{0} + {10.}^{n} C_{1} + . . . + 10^{n}$ $^{n} C_{n})$
$= 10 (n \sum r = 1$ $^{n} C_{r} (10)^{r} (1)^{n - r})$
$= 10 (1 + 10)^{n} = {10.11}^{n} = (B)$

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