1. Given sums

We need to evaluate

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

and

$$\beta=\sum_{k=0}^{n-1} \frac{\binom{n}{k}\binom{n}{k+1}}{k+2}.$$

Given that $5\alpha=6\beta,$ we must find $n$.

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2. Simplify $\alpha$

Use the identity

$$\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}.$$

So,

$$\alpha=\sum_{k=0}^n \binom{n}{k}\cdot \frac{1}{k+1}\binom{n}{k} =\frac{1}{n+1}\sum_{k=0}^n \binom{n}{k}\binom{n+1}{k+1}.$$

Now use

$$\binom{n+1}{k+1}=\binom{n+1}{n-k},$$

so

$$\alpha=\frac{1}{n+1}\sum_{k=0}^n \binom{n}{k}\binom{n+1}{n-k}.$$

By Vandermonde's identity,

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

Hence

$$\boxed{\alpha=\frac{1}{n+1}\binom{2n+1}{n}}.$$

Using

$$\binom{2n+1}{n}=\frac{2n+1}{n+1}\binom{2n}{n},$$

we also get

$$\alpha=\frac{1}{n+1}\cdot \frac{2n+1}{n+1}\binom{2n}{n} =\boxed{\frac{2n+1}{(n+1)^2}\binom{2n}{n}}.$$

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3. Simplify $\beta$

We use

$$\binom{n}{k+1}=\frac{n-k}{k+1}\binom{n}{k}.$$

So,

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

This form is not the easiest for summation. Instead rewrite directly as

$$\frac{1}{k+2}\binom{n}{k+1}=\frac{1}{(n+1)(n+2)}(k+1)\binom{n+2}{k+2}?$$

But an even cleaner route is:

First write

$$\frac{1}{k+2}\binom{n}{k+1}=\frac{1}{n+1}\cdot \frac{1}{n+2}\,(k+1)\binom{n+2}{k+2},$$

which is cumbersome.

A better substitution is to shift index. Let $r=k+1$. Then $r=1$ to $n$, and

$$\beta=\sum_{r=1}^{n}\frac{\binom{n}{r-1}\binom{n}{r}}{r+1}.$$

Now use

$$\binom{n}{r-1}=\frac{r}{n-r+1}\binom{n}{r},$$

but again this is messy.

So let us instead convert one factor using

$$\frac{1}{k+2}\binom{n}{k+1} = \frac{1}{n+1}\binom{n+1}{k+2}.$$

Indeed,

$$\binom{n+1}{k+2}=\frac{(n+1)!}{(k+2)!(n-k-1)!} =\frac{n+1}{k+2}\binom{n}{k+1}.$$

Hence

$$\frac{1}{k+2}\binom{n}{k+1}=\frac{1}{n+1}\binom{n+1}{k+2}.$$

Therefore,

$$\beta=\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n}{k}\binom{n+1}{k+2}.$$

Now use symmetry:

$$\binom{n+1}{k+2}=\binom{n+1}{n-k-1}.$$

So

$$\beta=\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n}{k}\binom{n+1}{n-k-1}.$$

Applying Vandermonde,

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

Thus

$$\boxed{\beta=\frac{1}{n+1}\binom{2n+1}{n-1}}.$$

Now express it in terms of $\binom{2n}{n}$:

$$\binom{2n+1}{n-1}=\frac{(2n+1)!}{(n-1)!(n+2)!}.$$

Also,

$$\binom{2n}{n}=\frac{(2n)!}{n!n!}.$$

So

$$\binom{2n+1}{n-1}=\binom{2n}{n}\cdot \frac{(2n+1)n}{(n+1)(n+2)}.$$

Hence

$$\beta=\frac{1}{n+1}\cdot \binom{2n}{n}\cdot \frac{(2n+1)n}{(n+1)(n+2)} =\boxed{\frac{n(2n+1)}{(n+1)^2(n+2)}\binom{2n}{n}}.$$

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4. Use the condition $5\alpha=6\beta$

Substitute the expressions for $\alpha$ and $\beta$:

$$5\cdot \frac{2n+1}{(n+1)^2}\binom{2n}{n} = 6\cdot \frac{n(2n+1)}{(n+1)^2(n+2)}\binom{2n}{n}.$$

Cancel the common nonzero factor

$$\frac{2n+1}{(n+1)^2}\binom{2n}{n}.$$

We get

$$5=\frac{6n}{n+2}.$$

So,

$$5(n+2)=6n$$ $$5n+10=6n$$ $$\boxed{n=10}.$$

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5. Comparison with stored answer

Stored correct answer: $10$

Our derived answer is also $10$, so it agrees.