The coefficient of x^{100} in the expansion o | vedclass

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(D) The given expression is a geometric series: $S = \sum_{j=0}^{200} (1+x)^j = 1 + (1+x) + (1+x)^2 + \dots + (1+x)^{200}$.This is a geometric progres

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$\binom{201}{100}$

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(D) The given expression is a geometric series: $S = \sum_{j=0}^{200} (1+x)^j = 1 + (1+x) + (1+x)^2 + \dots + (1+x)^{200}$.This is a geometric progression with first term $a = 1$,common ratio $r = (1+x)$,and number of terms $n = 201$.The sum is given by $S = \frac{a(r^n - 1)}{r - 1} = \frac{1((1+x)^{201} - 1)}{(1+x) - 1} = \frac{(1+x)^{201} - 1}{x}$.We need the coefficient of $x^{100}$ in $S = \frac{(1+x)^{201} - 1}{x}$.This is equivalent to finding the coefficient of $x^{101}$ in the expansion of $(1+x)^{201} - 1$.The general term in the expansion of $(1+x)^{201}$ is $\binom{201}{k} x^k$.For $k = 101$,the coefficient is $\binom{201}{101}$.Note: $\binom{201}{101} = \binom{201}{201-101} = \binom{201}{100}$.Thus,the coefficient of $x^{100}$ is $\binom{201}{100}$.

The coefficient of $x^{100}$ in the expansion of $\sum_{j=0}^{200} (1 + x)^j$ is

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