If the coefficient of $$x100$$ in $$1+(1+x)+(1+x)2+$$…$$+(1+x)n$$;(n$$≥$$100) is $$201C101$$ , then n is equal to

$$1+(1+x)+(1+x)2+$$…$$+(1+x)n$$ $$=(1+x)n+1-11+x-1=(1+x)n+1-1x$$ the coefficientof $$x100$$ in $$(1+x)n+1-1x=the$$ coefficientof $$x101$$ in $$(1+x)n+1-1=n+1C101=201C101$$ ⇒$$n=200$$

Binomial theorem for positive integral Index

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Question

$If the coefficient of x^{100} in 1 + (1 + x) + (1 + x)^{2} + \dots + (1 + x)^{n}; (n \geq 100) {is}^{201} C_{101}, then n is equal to$

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Solution

$1 + (1 + x) + (1 + x)^{2} + \dots + (1 + x)^{n} = \frac{(1 + x)^{n + 1} - 1}{1 + x - 1} = \frac{(1 + x)^{n + 1} - 1}{x} the coefficient of x^{100} in \frac{(1 + x)^{n + 1} - 1}{x} = the coefficientof x^{101} in (1 + x)^{n + 1} - 1 =^{n + 1} C_{101} =^{201} C_{101} \Rightarrow n = 200$

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