If the sum that you're interested in is: $$ \sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}}, $$ Then, $$ \sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}}=\sum_{k=1}^{50}\frac{k^2(n-k+1)}{k}=\sum_{k=1}^{50}k(n-k+1)=(n+1)\sum_{k=1}^{50}k-\sum_{k=1}^{50}k^2. $$

If the sum that you're interested in is: $$ \sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}}, $$ Then, $$ \sum_{k=1}^{50}\frac{k^2\binom{n}{k}}{\binom{n}{k-1}}=\sum_{k=1}^{50}\frac{k^2(n-k+1)}{k}=\sum_{k=1}^{50}k(n-k+1)=(n+1)\sum_{k=1}^{50}k-\sum_{k=1}^{50}k^2. $$ Then, using the formulas $$ \sum_{k=1}^{50}k=\frac{50\cdot 51}{2}=1275 $$ and $$ \sum_{k=1}^{50}k^2=\frac{50\cdot 51\cdot 101}{6}=42925, $$ you will get the negative of the answer at the end of the post (perhaps due to the negative confusion in the discussion above).