There is a hint for this ex.: using symmetry. I would appreciate another hint to take advantage of.

My approach so far:

Induction step:

$\sum_{k=0}^{n+1} \binom{n+1}{k}^2$

$= 1 + \sum_{k=1}^{n+1} \binom{n+1}{k}^2$

$= 1 + \sum_{k=0}^{n} \binom{n+1}{k+1}^2$

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

$= 1 + 2\cdot\sum_{k=0}^{n} (\binom{n}{k}\binom{n}{k+1}) + \sum_{k=0}^{n} \binom{n}{k}^2 + \sum_{k=0}^{n} \binom{n}{k+1}^2 $

Now, I should simplify $2\sum_{k=0}^{n} (\binom{n}{k}\binom{n}{k+1})$ but I don't know how to proceed.


marked as duplicate by Servaes, Arthur, user10354138, MathOverview, N. F. Taussig Nov 4 '18 at 17:17

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  • $\begingroup$ $\binom{n}{k}^2=\binom{n}{k}\cdot\binom{n}{n-k}$ $\endgroup$ – Tito Eliatron Nov 4 '18 at 16:58
  • $\begingroup$ Do you know an answer that uses induction? $\endgroup$ – RM777 Nov 4 '18 at 17:15

There is a great combinatoric interpretation of this problem! Consider a lattice grid. $\binom{2n}{n}$ is the number of ways to get from $(0,0)$ to $(n,n)$ by only taking steps to the rihgt and up: this is because a total of $2n$ steps must be taken, and $n$ of them must be "chosen" to be up, the rest will be to the right.

Now $\binom{n}{k}$ is the number of ways to get from $(0,0)$ to $(n-k,k)$; of $n$ steps, you select $k$ to be up, the rest ($n-k$) will be to the right.

If you want to continue from $(n-k,k)$ to $(n,n)$, there are $n$ steps left in your journey (we've taken $n$, and $2n$ total must be taken). Note that $k$ of these steps must be to the right, and the result will be up. There are $\binom{n}{k}$ ways to make this decision.

Thus there are $\binom{n}{k}$ ways to get to $(n-k,k)$ from the origin then $\binom{n}{k}$ ways to get from there to $(n,n)$. These selections are independent so the total number of paths to $(n,n)$ that pass through $(n-k,k)$ is $\binom{n}{k}^2$.

Every path to $(n,n)$ must pass through $(n-k,k)$ for exactly one value of $k$, so the desired result is gotten by summing over $k$.

Edit: the symmetry used is that the square with vertices $(0,0),(0,n),(n,0),(n,n)$ is symmetric about the line $(a,b)$ for $a+b=n$

  • $\begingroup$ Thx I thought the only proof is the induction proof. Your expl. makes sense alright but does this verify the truth of your args? $\endgroup$ – RM777 Nov 4 '18 at 20:24

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