# Combinatorics - Binomial Theorem Questions

1. Oct 14, 2006

### mattmns

There are a few questions that have been giving me trouble with this binomial theorem stuff.

(1). Using the binomial theorem and the relation $$(1+x)^{m_1} (1+x)^{m_2} = (1+x)^{m_1 + m_2}$$

prove that:

$$\sum_{k=0}^n \binom{m_1}{k} \binom{m_2}{n-k} = \binom{m_1 + m_2}{n}$$

(2). Prove by induction on n that, for n a positive integer,

$$\frac{1}{(1-z)^n} = \sum_{k=0}^\infty \binom{n+k-1}{k}z^k, |z| < 1.$$

Assume the validity of

$$\frac{1}{(1-z)} = \sum_{k=0}^\infty z^k, |z| < 1.$$

-----------------------

For (1). This is very easy to prove using a combinatorial argument, but I am just not seeing how I can prove it with the binomial theorem. I have been pluggin them in getting the sums, but nothing is clicking. This problem may be similar to the problem with the other problem (mixing of sums?)

For (2). The base case is obvious, practically assumed. But I am not sure where to go with the following:

After some basic maniuplation I get:

$$\frac{1}{(1-z)^{n+1}} = \left(\sum_{k=0}^\infty \binom{n+k-1}{k}z^k\right)\left(\sum_{1=0}^\infty z^i \right)$$

Can I mix these two together somehow?

My goal is to get the above equation equal to $$\sum_{k=0}^\infty \binom{n+k}{k}z^k\right)\left$$

Any hints or ideas on either of the problems? Thanks!

Last edited: Oct 14, 2006
2. Oct 15, 2006

### Galileo

In both cases you have to multiply sums. If you multiply two polynomials, you get another polynomial:

$$\sum \limits_{k=0}^{m_1} a_kx^k\cdot\sum \limits_{k=0}^{m_2} b_kx^k=\sum \limits_{k=0}^{m_1+m_2} c_kx^k$$
First find out how c_k is related to the coefficients a_i and b_i. After that, it's plug and play for both exercises.