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?

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$\frac{1}{1+x}\centerdot(1+x)^{2n} = (1+x)^n$ iff $n=1$; are you sure that you wrote down the right equation? – Brian M. Scott Jun 14 '12 at 10:59
no, sorry. i corrected it... – EMDB1 Jun 14 '12 at 11:01

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 $$\sum_{i=0}^k(-1)^i\binom{n-1+i}i\binom{2n}{k-i}\;.$$

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WHOA! How do you do the mouse-over effect?! – AlanH Mar 24 '13 at 7:22
@Alan: You can only get one paragraph of it. You get that by starting with >!. – Brian M. Scott Mar 24 '13 at 18:17

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.

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