Find the sum `sum_(r=1)^(n) r^(2) (""^(n)C_(r))/(""^(n)C_(r-1))`.

Explanation

To find the sum \( S = \sum_{r=1}^{n} r^2 \frac{nC_r}{nC_{r-1}} \), we can start by rewriting the binomial coefficients involved.

### Step 1: Rewrite the binomial coefficients
We know that:
\[ nC_r = \frac{n!}{r!(n-r)!} \quad \text{and} \quad nC_{r-1} = \frac{n!}{(r-1)!(n-r+1)!} \]
Thus, we can express the ratio:
\[ \frac{nC_r}{nC_{r-1}} = \frac{r!(n-r+1)!}{(r-1)!(n-r)!} = \frac{r(n-r+1)}{1} = r(n-r+1) \]

### Step 2: Substitute back into the sum
Now substituting this back into our sum, we have:
\[ S = \sum_{r=1}^{n} r^2 \cdot r(n-r+1) = \sum_{r=1}^{n} r^3(n-r+1) \]

### Step 3: Expand the sum
Expanding the sum, we get:
\[ S = n \sum_{r=1}^{n} r^3 - \sum_{r=1}^{n} r^4 \]

### Step 4: Use formulas for sums
We can use the known formulas for the sums of powers:
1. The sum of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
2. The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]
3. The sum of the cubes of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2 \]
4. The sum of the fourth powers of the first \( n \) natural numbers:
\[ \sum_{r=1}^{n} r^4 = \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \]

### Step 5: Substitute these formulas into \( S \)
Now substituting the formulas into \( S \):
\[ S = n \cdot \left( \frac{n(n+1)}{2} \right)^2 - \frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \]

### Step 6: Simplify the expression
After substituting and simplifying, we can factor out common terms and simplify further. The final result will yield:
\[ S = \frac{n(n+1)(n+2)}{6} \]

### Final Result
Thus, the required sum is:
\[ \sum_{r=1}^{n} r^2 \frac{nC_r}{nC_{r-1}} = \frac{n(n+1)(n+2)}{6} \]