If n geq 100 and the coefficient of x^{100} i | vedclass

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(B) The given expression is a geometric series: $S = 1 + (1+x) + (1+x)^2 + \cdots + (1+x)^n$. Using the formula for the sum of a geometric series $S =

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$200$

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(B) The given expression is a geometric series: $S = 1 + (1+x) + (1+x)^2 + \cdots + (1+x)^n$. Using the formula for the sum of a geometric series $S = \frac{a(r^n - 1)}{r - 1}$,where $a = 1$,$r = (1+x)$,and the number of terms is $n+1$: $S = \frac{1((1+x)^{n+1} - 1)}{(1+x) - 1} = \frac{(1+x)^{n+1} - 1}{x}$. We need the coefficient of $x^{100}$ in $S$,which is the coefficient of $x^{100}$ in $\frac{(1+x)^{n+1} - 1}{x}$. This is equivalent to finding the coefficient of $x^{101}$ in $(1+x)^{n+1} - 1$. The coefficient of $x^{101}$ in $(1+x)^{n+1}$ is given by ${ }^{n+1} C_{101}$. Given that this coefficient is ${ }^{201} C_{101}$,we have ${ }^{n+1} C_{101} = { }^{201} C_{101}$. Therefore,$n+1 = 201$,which implies $n = 200$.

If $n \geq 100$ and the coefficient of $x^{100}$ in $1+(1+x)+(1+x)^2+\cdots+(1+x)^n$ is ${ }^{201} C_{101}$,then $n=$

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