Let $$\alpha = \sum_{k=0}^{n} \frac{{}^nC_k^2}{k+1}$$ and $$\beta = \sum_{k=0}^{n-1} \frac{{}^nC_k \cdot {}^nC_{k+1}}{k+2}$$. If $$5\alpha = 6\beta$$, then $$n$$ equals

We start with the two sums
$$\alpha = \sum_{k=0}^{n} \frac{\binom{n}{k}^{2}}{k+1},\qquad \beta = \sum_{k=0}^{\,n-1} \frac{\binom{n}{k}\,\binom{n}{k+1}}{k+2}.$$

Case 1: Closed form for $$\alpha$$
First use the relation
$$\frac{\binom{n}{k}}{k+1}=\frac{1}{\,n+1\,}\binom{\,n+1\,}{\,k+1\,}.$$ Hence

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

Notice that $$\binom{\,n+1\,}{\,k+1\,}=\binom{\,n+1\,}{\,n-k\,}.$$
Putting $$j=n-k$$ the summation becomes $$ \alpha=\frac{1}{n+1}\sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,}. $$

By Vandermonde’s identity
$$ \sum_{j=0}^{n}\binom{n}{j}\binom{\,n+1\,}{\,j\,} =\binom{\,2n+1\,}{\,n\,}. $$ Therefore

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

Case 2: Closed form for $$\beta$$

Rewrite the general term of $$\beta$$ with a shifted index $$r=k+1$$ ($$1\le r\le n$$):
$$ \beta=\sum_{r=1}^{n}\frac{\binom{n}{r-1}\,\binom{n}{r}}{r+1}. $$

Use $$ (r+1)\binom{n}{r}=n\binom{\,n-1\,}{\,r-1\,}. $$ Multiplying numerator and denominator of each term by $$r+1$$ gives

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

Put $$s=r-1$$ (so $$0\le s\le n-1$$):
$$ \beta=\frac{n}{(n+1)(n+2)} \sum_{s=0}^{n-1}\binom{\,n-1\,}{\,s\,}\binom{\,n+2\,}{\,s+2\,}. $$

Since $$\binom{\,n+2\,}{\,s+2\,}=\binom{\,n+2\,}{\,n-s\,},$$ the sum equals the coefficient of $$x^{\,n}$$ in $$(1+x)^{\,n-1}(1+x)^{\,n+2}=(1+x)^{\,2n+1},$$ which by the binomial theorem is $$\binom{\,2n+1\,}{\,n\,}.$$ Hence

$$ \boxed{\beta=\dfrac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}.} $$

Case 3: Using the given relation
The problem states $$5\alpha = 6\beta.$$ Insert the closed forms:

$$ 5\left[\frac{1}{n+1}\binom{\,2n+1\,}{\,n\,}\right] =6\left[\frac{n}{(n+1)(n+2)}\binom{\,2n+1\,}{\,n\,}\right]. $$

Because $$\binom{\,2n+1\,}{\,n\,}\neq 0$$, it cancels out, and so does the common factor $$n+1$$, giving

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

Solve for $$n$$:
$$ 5(n+2)=6n \;\;\Longrightarrow\;\; 5n+10=6n \;\;\Longrightarrow\;\; n=10. $$

Hence $$n=10$$.