Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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


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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.