Binomial TheoremHard

Question

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x(1 + x)^{999} + x^{2}(1 + x)^{998} + \ldots\ldots + x^{1000}$ is

Options

A.$\ ^{1001}C_{501}$
B.$\ ^{1002}C_{500}$
C.$\ ^{1002}C_{501}$
D.$\ ^{1000}C_{501}$

Solution

$S = (1 + x)^{1000} + x(1 + x)^{999} + x^{2}(1 + x)^{998} + \ldots. + x^{1000}$

$${= (1 + x)^{1000}\frac{\left( 1 - \left( \frac{x}{1 + x} \right)^{1001} \right)}{1 - \frac{x}{1 + x}} }{= (1 + x)^{1001} - x^{1001} }$$Required sum $= \ ^{1001}C_{499} + \ ^{1001}C_{500} = \ ^{1002}C_{500}$

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