How can it be shown that $\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}^{2}a^{n-k}$

Question

How can it be shown that:

$$\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}^{2}a^{n-k}$$

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My try:

$$\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\left(a-1\right)^k=\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{n}\sum_{j=0}^{k}\binom{k}{j}a^{j}\left(-1\right)^{k-j}$$$$=\left(-1\right)^{n}\sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{n-k}\sum_{j=0}^{k}\binom{k}{j}a^{j}\left(-1\right)^{-j}$$ $$=\left(-1\right)^{n}\sum_{j=0}^{n}\binom{n}{j}a^{j}\left(-1\right)^{-j}\sum_{k=j}^{n}\binom{n-j}{k-j}\binom{-n-1}{n-k}$$

$$=\left(-1\right)^{n}\sum_{j=0}^{n}\binom{n}{j}a^{j}\left(-1\right)^{-j}\binom{-j-1}{n-j}$$

$$=\sum_{j=0}^{n}\binom{n}{j}a^{j}\binom{n}{j}=\sum_{\color{red}{j}=0}^{n}\binom{n}{\color{red}{j}}^2a^{n-\color{red}{j}}$$

The problem is that I have $\color{red}{j}$ instead of $k$.

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Source :math.wvu.edu

Answer 1

Replace $j$ by $k$ to get the desired result. Calling the summation index $j$ or $k$ does not make a difference.

Answer 2

Here is a slightly different approach, for variety. We seek to show that

$$\sum_{k=0}^n {n\choose k} {2n-k\choose n} (a-1)^k = \sum_{k=0}^n {n\choose k}^2 a^{n-k}.$$

Note that with $0\le q\le n$ the coefficients on $[a^q]$ must be equal, hence we require

$$\sum_{k=q}^n {n\choose k} {2n-k\choose n} {k\choose q} (-1)^{k-q} = {n\choose q}^2.$$

Starting with the left we note that

$${n\choose k} {k\choose q} = \frac{n!}{(n-k)! \times q! \times (k-q)!} = {n\choose q} {n-q\choose n-k}.$$

We are now reduced to showing that

$$\sum_{k=q}^n {2n-k\choose n} {n-q\choose n-k} (-1)^{k-q} = {n\choose q}.$$

Starting with the LHS we find

$$\sum_{k=q}^n {2n-k\choose n-k} {n-q\choose n-k} (-1)^{k-q} \\ = [z^n] (1+z)^{2n} \sum_{k=q}^n \frac{z^k}{(1+z)^k} {n-q\choose n-k} (-1)^{k-q} \\ = [z^n] (1+z)^{2n} \frac{z^q}{(1+z)^q} \sum_{k=0}^{n-q} \frac{z^k}{(1+z)^k} {n-q\choose n-q-k} (-1)^{k} \\ = [z^n] (1+z)^{2n} \frac{z^q}{(1+z)^q} \left(1-\frac{z}{1+z}\right)^{n-q} \\ = [z^n] (1+z)^{2n} \frac{z^q}{(1+z)^q} \frac{1}{(1+z)^{n-q}} = [z^{n-q}] (1+z)^n = {n\choose q}.$$

We obtain the RHS and this concludes the argument.