The general term of the series is $\sum_{r=0}^{100}\binom{500}{r}\binom{500-r}{400}2^{100-r}$

What I tried to think that this series was an expansion for a series inside a series but the thing that i found out is that I have to find the sum of a series which stops at 100 but runs till the power of 500 .

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    $\begingroup$ Use $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ and play around a bit to find $\binom{100}{r}$. $\endgroup$ – Daniel Fischer Mar 11 '16 at 16:48

It is equal to $\sum_{r=0}^{100}\frac{500!}{r!(500-r)!}\frac{(500-r)!}{400!(100-r)!}2^{100-r}=\frac{500!}{400!100!}\sum_{r=0}^{100}\frac{100!}{r!(100-r)!}2^{100-r}={500\choose 100}3^{100}$.

  • $\begingroup$ All too easy. +1. $\endgroup$ – Mark Viola Mar 11 '16 at 17:15

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