The coefficient of x^{301} in (1+x)^{500} + x | vedclass

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(D) The given expression is a geometric series with first term $a = (1+x)^{500}$,common ratio $r = \frac{x}{1+x}$,and number of terms $n = 501$.The su

Vedclass · Accepted Answer

$^{501}C_{200}$

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(D) The given expression is a geometric series with first term $a = (1+x)^{500}$,common ratio $r = \frac{x}{1+x}$,and number of terms $n = 501$.The sum of the series is $S = a \frac{1-r^n}{1-r} = (1+x)^{500} \left[ \frac{1 - (\frac{x}{1+x})^{501}}{1 - \frac{x}{1+x}} \right]$.Simplifying the expression:$S = (1+x)^{500} \left[ \frac{\frac{(1+x)^{501} - x^{501}}{(1+x)^{501}}}{\frac{1+x-x}{1+x}} \right] = (1+x)^{500} \cdot \frac{(1+x)^{501} - x^{501}}{(1+x)^{501}} \cdot (1+x) = (1+x)^{501} - x^{501}$.We need the coefficient of $x^{301}$ in $(1+x)^{501} - x^{501}$.The coefficient of $x^{301}$ in $(1+x)^{501}$ is $^{501}C_{301}$.Using the property $^{n}C_{r} = ^{n}C_{n-r}$,we have $^{501}C_{301} = ^{501}C_{501-301} = ^{501}C_{200}$.

The coefficient of $x^{301}$ in $(1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \ldots + x^{500}$ is:

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