The coefficients of $x^{301}$ in the expansion of ${(1 + x)}^{500} + {x (1 + x)}^{499} + x^{2} {(1 + x)}^{498} + . . . + x^{500}$ is

^{501} C_{301}

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^{500} C_{301}

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^{501} C_{300}

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none of these

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Solution

The correct option is C $^{501} C_{301}$
The given series
$S = {(1 + x)}^{500} [1 + \frac{x}{1 + x} + {(\frac{x}{1 + x})}^{2} + . . . + {(\frac{x}{1 + x})}^{500}]$

$= {(1 + x)}^{500} \times \frac{1 - {(\frac{x}{1 + x})}^{501}}{1 - \frac{x}{1 + x}} = {(1 + x)}^{501} - x^{501}$

Hence, the coefficient of $x^{301}$ in $S =^{501} C_{301} .$

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