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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

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    no, sorry. i corrected it...2012-06-14

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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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    @Alan: You can only get one paragraph of it. You get that by starting with `>!`.2013-03-24
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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.