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What is formula for multiple two finite series?
$$\sum_{k=0}^m {m \choose k}x^k \sum_{k=0}^n {n \choose k}x^k$$

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up vote 8 down vote accepted

The finite series are just polynomials, so you multiply them as such. More generally, the Cauchy product of two formal power series is obtained by multiplying them as if they were polynomials. Thus, the coefficient of $x^n$ in

$$\left(\sum_{k\ge 0}a_kx^k\right)\left(\sum_{k\ge 0}b_kx^k\right)$$

is $$\sum_{k=0}^na_kb_{n-k}\;.$$

In your problem, if

$$\left(\sum_{k=0}^m\binom{m}kx^k\right)\left(\sum_{k=0}^n\binom{n}kx^k\right)=\sum_{k=0}^{m+n}c_kx^k\;,$$

then $$c_k=\sum_{i=0}^k\binom{m}i\binom{n}{k-i}=\binom{m+n}k\;.$$

Added: That last step uses Vandermonde’s identity, which is easily proved either from the binomial theorem, as in lab bhattacharjee’s answer, or by a purely combinatorial argument.

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Using well known Binomial Theorem, $(a+b)^n=\sum_{k=0}^n {n \choose k}a^{n-k}b^k$ for natural $n,$

$$\left(\sum_{k=0}^m {m \choose k}x^k \right) \left(\sum_{k=0}^n {n \choose k}x^k\right)=(1+x)^m(1+x)^n=(1+x)^{m+n}=\sum_{r=0}^{m+n} {{m+n} \choose r}x^r$$

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