The coefficient of x^{50} in the binomial exp | vedclass

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(D) The given expression is a geometric series with first term $a = (1 + x)^{1000}$,common ratio $r = \frac{x}{1 + x}$,and $n = 1001$ terms.Using the

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$\frac{1001!}{50!951!}$

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(D) The given expression is a geometric series with first term $a = (1 + x)^{1000}$,common ratio $r = \frac{x}{1 + x}$,and $n = 1001$ terms.Using the sum formula for a geometric series $S = \frac{a(1 - r^n)}{1 - r}$:$S = \frac{(1 + x)^{1000} \left[ 1 - \left( \frac{x}{1 + x} \right)^{1001} \right]}{1 - \frac{x}{1 + x}}$$S = \frac{(1 + x)^{1000} \left[ \frac{(1 + x)^{1001} - x^{1001}}{(1 + x)^{1001}} \right]}{\frac{1 + x - x}{1 + x}}$$S = \frac{(1 + x)^{1000} \left[ (1 + x)^{1001} - x^{1001} \right]}{(1 + x)^{1001} \cdot \frac{1}{1 + x}}$$S = (1 + x)^{1001} - x^{1001}$We need the coefficient of $x^{50}$ in $(1 + x)^{1001} - x^{1001}$.Since $x^{1001}$ does not contain $x^{50}$,the coefficient is the same as the coefficient of $x^{50}$ in $(1 + x)^{1001}$,which is given by $^{1001}C_{50} = \frac{1001!}{50!951!}$.

The coefficient of $x^{50}$ in the binomial expansion of $(1 + x)^{1000} + x(1 + x)^{999} + x^{2}(1 + x)^{998} + \dots + x^{1000}$ is

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