Question:

Step 1: Apply Vandermonde's Identity to simplify $S_n$. The sum is given by $S_n = \sum_{r=0}^{n-1}\binom{n}{r}\binom{n}{r+1}$. Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite $\binom{n}{r+1}$ as $\binom{n}{n-(r+1)}$. Thus, $S_n = \sum_{r=0}^{n-1}\binom{n}{r}\binom{n}{n-r-1}$. By Vandermonde's Identity, $\sum_{k=0}^{r} \binom{m}{k}\binom{p}{r-k} = \binom{m+p}{r}$. Setting $m=n$, $p=n$, and $r=n-1$, we obtain: $$S_n = \binom{n+n}{n-1} = \binom{2n}{n-1}$$ Step 2: Determine the expression for $\frac{S_{n+1}}{S_n}$. Using the identity from Step 1, we find $S_{n+1}$: $$S_{n+1} = \binom{2(n+1)}{(n+1)-1} = \binom{2n+2}{n}$$ Now, we compute the ratio $\frac{S_{n+1}}{S_n}$: $$\frac{S_{n+1}}{S_n} = \frac{\binom{2n+2}{n}}{\binom{2n}{n-1}}$$ Expanding the binomial coefficients: $$ \frac{S_{n+1}}{S_n} = \frac{\frac{(2n+2)!}{n!(2n+2-n)!}}{\frac{(2n)!}{(n-1)!(2n-(n-1))!}} = \frac{\frac{(2n+2)!}{n!(n+2)!}}{\frac{(2n)!}{(n-1)!(n+1)!}} $$ $$ = \frac{(2n+2)!}{n!(n+2)!} \cdot \frac{(n-1)!(n+1)!}{(2n)!} $$ $$ = \frac{(2n+2)(2n+1)(2n)!}{n(n-1)!(n+2)(n+1)!} \cdot \frac{(n-1)!(n+1)!}{(2n)!} $$ $$ = \frac{(2n+2)(2n+1)}{n(n+2)} $$ Step 3: Set up the equation for $n$. We are given that $\frac{S_{n+1}}{S_n} = \frac{15}{4}$. $$ \frac{(2n+2)(2n+1)}{n(n+2)} = \frac{15}{4} $$ Step 4: Solve the equation for $n$. $$ \frac{2(n+1)(2n+1)}{n(n+2)} = \frac{15}{4} $$ $$ 8(n+1)(2n+1) = 15n(n+2) $$ $$ 8(2n^2 + 3n + 1) = 15n^2 + 30n $$ $$ 16n^2 + 24n + 8 = 15n^2 + 30n $$ $$ n^2 - 6n + 8 = 0 $$ Factoring the quadratic equation: $$ (n-2)(n-4) = 0 $$ This yields two possible values for $n$: $n=2$ or $n=4$. Step 5: Find the sum of all possible values of $n$. Since $n \in \mathbb{N}$, both $n=2$ and $n=4$ are valid natural numbers. The sum of all possible values of $n$ is $2+4=6$.