Question about generating functions I have question about generating functions.
I need to make this equation:
$(\frac{1}{1+x})^n\centerdot(1+x)^{2n} = (1+x)^n$
in this form:
$\sum\limits_{i=0}^{k}(-1)^iD(?,?)\binom{?}{?} = \binom{n}{k}$
How can I do this?
Thanks in advance
 A: The coefficient of $x^k$ in $(1+x)^n$ is $\binom{n}k$, so you want to work out the coefficient of $x^k$ on the lefthand side. You know that $$(1+x)^{2n}=\sum_{i\ge o}\binom{2n}ix^i\;,$$ and you probably know that $$\frac1{(1-x)^n}=\sum_{i\ge 0}\binom{n-1+i}ix^i\;,$$ so that $$\frac1{(1+x)^n}=\sum_{i\ge 0}(-1)^i\binom{n-1+i}ix^i\;.$$
Thus, $$\frac{(1+x)^{2n}}{(1+x)^n}=\left(\sum_{i\ge 0}(-1)^i\binom{n-1+i}ix^i\right)\left(\sum_{i\ge 0}\binom{2n}ix^i\right)\;.\tag{1}$$
Now just expand to find the coefficient of $x^k$. I’ve done the rest below but spoiler-protected it; mouse-over to see it.

 The coefficient of $x^k$ in the product $$\left(\sum_{i\ge 0}a_ix^i\right)\left(\sum_{i\ge 0}b_ix^i\right)$$ is $$\sum_{i=0}^ka_ib_{k-i}\;,$$ so the coefficient of $x^k$ in $(1)$ is $$\sum_{i=0}^k(-1)^i\binom{n-1+i}i\binom{2n}{k-i}\;.$$

A: Expand $(1+x)^{-n}$ from the left hand side using the binomial series and $(1+x)^{2n}$ using the ordinary binomial formula; ditto on the right hand side. Multiply the factors on the left, collect equal powers of $x$ and compare coefficients.
