1. We need to evaluate $S=\sum_{r=1}^{30} \frac{r^2\binom{30}{r}^2}{\binom{30}{r-1}}.$ We are given that $S=\alpha\cdot 2^{29}$ and we must find $\alpha$.

2. First simplify the summand. Use $\binom{30}{r}=\frac{30!}{r!(30-r)!},\qquad \binom{30}{r-1}=\frac{30!}{(r-1)!(31-r)!}.$ Then $\frac{\binom{30}{r}}{\binom{30}{r-1}}=\frac{30-r+1}{r}=\frac{31-r}{r}.$ So,

Hence, $\frac{r^2\binom{30}{r}^2}{\binom{30}{r-1}}=r(31-r)\binom{30}{r}.$ Therefore $S=\sum_{r=1}^{30} r(31-r)\binom{30}{r}.$

3. Expand: $r(31-r)=31r-r^2.$ Thus $S=31\sum_{r=1}^{30} r\binom{30}{r}-\sum_{r=1}^{30} r^2\binom{30}{r}.$ So we need the standard sums $\sum_{r=0}^{n} r\binom{n}{r}=n2^{n-1},$ $\sum_{r=0}^{n} r(r-1)\binom{n}{r}=n(n-1)2^{n-2}.$ Also, $r^2=r(r-1)+r,$ so

4. Put $n=30$. Then $\sum_{r=0}^{30} r\binom{30}{r}=30\cdot 2^{29},$ and $\sum_{r=0}^{30} r^2\binom{30}{r}=30\cdot 29\cdot 2^{28}+30\cdot 2^{29}.$ Since the $r=0$ term is $0$, the same formulas apply for summation from $1$ to $30$.

5. Substitute into $S$: $S=31(30\cdot 2^{29})-\left(30\cdot 29\cdot 2^{28}+30\cdot 2^{29}\right).$ Write everything in terms of $2^{28}$: $31\cdot 30\cdot 2^{29}=31\cdot 30\cdot 2\cdot 2^{28}=1860\cdot 2^{28},$ $30\cdot 29\cdot 2^{28}=870\cdot 2^{28},$ $30\cdot 2^{29}=60\cdot 2^{28}.$ Therefore $S=(1860-870-60)2^{28}=930\cdot 2^{28}.$ Now, $930\cdot 2^{28}=465\cdot 2^{29}.$ Hence $\alpha=465.$

6. Comparison with stored answer: Stored correct answer = $465$. Our derived answer also equals $465$, so they agree.