<s>[INST] How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$ How prove this  $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$
This equation How prove it? Thank you
I want take this  $$f(x)=(1-x)^{n+1}?$$
But I can't deal this $[n/3]$,
Thank you for you help
 [/INST] Start from the RHS. We are counting the number of ways to choice $k$ elements among $n+1$, then choice $n-k$ elements among the $k$ elements previously chosen. If we imagine to assign a $+1$ weight in the first step, then increase the weight in the second step, we are counting the number of ways to assign a $+1$ or $+2$ weigth to the elements of a subset of $\{1,\ldots,n+1\}$ in such a way that the sum of the weights is just $n$. Rephrasing in the analytic combinatorics framework:
$$RHS=\sum_{k=\lfloor n/2\rfloor}^{n}\binom{n+1}{k}\binom{k}{n-k}=[x^n]\,\left((1+x+x^2)^{n+1}\right).\tag{1}$$
Now $1+x+x^2  =(1+x)^2-x$, so, for istance:
$$[x^n](1+x+x^2)^{n+1} = [x^n]\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^j x^j(x+1)^{2n+2-2j}=\sum_{j=0}^{n+1}(-1)^j\binom{n+1}{j}\binom{2n+2-2j}{n-j},$$
just like $1+x+x^2=\frac{1-x^3}{1-x}$ and
$$\frac{1}{(1-x)^m}=\sum_{j=0}^{+\infty}\binom{m+j-1}{j}x^j,$$
give:
$$[x^n](1+x+x^2)^{n+1}=[x^n]\left(\sum_{j=0}^{n+1}\binom{n+1}{j}(-1)^j x^{3j}\right)\cdot\left(\sum_{j=0}^{+\infty}\binom{n+j}{j}x^j\right).\tag{2}$$
Regarding $(2)$ as a Cauchy product gives $RHS=LHS$, QED.
The saddle-point method gives also:
$$[x^n](1+x+x^2)^{n+1}=\frac{3^{n+2}}{\sqrt{\pi(12n+30)}}\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right).$$
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