If $\displaystyle\sum_{r=1}^{30}\dfrac{r^{2}\binom{30}{r}^{2}}{\binom{30}{r-1}}=\alpha\times 2^{29}$, then $\alpha$ is equal to \rule{2cm}{0.4pt}.
Step-by-Step Solution
Key Concept: $\dfrac{\binom{30}{r}}{\binom{30}{r-1}}=\dfrac{31-r}{r}$, so $\dfrac{r^{2}\binom{30}{r}^{2}}{\binom{30}{r-1}}=r(31-r)\binom{30}{r}.$ Then $r\binom{30}{r}=30\binom{29}{r-1}$ converts the sum into one over $\binom{29}{s}.$
$\dfrac{r^{2}\binom{30}{r}^{2}}{\binom{30}{r-1}}=r\binom{30}{r}\cdot(31-r)=30\binom{29}{r-1}(31-r).$
Let $s=r-1$ so $r$ runs $1$–$30$ $\Leftrightarrow$ $s$ runs $0$–$29$, $31-r=30-s$:
$$\sum_{s=0}^{29}30(30-s)\binom{29}{s}=30\!\left[30\sum_{s=0}^{29}\binom{29}{s}-\sum_{s=0}^{29}s\binom{29}{s}\right].$$
$\sum\binom{29}{s}=2^{29},\ \sum s\binom{29}{s}=29\cdot 2^{28}.$ So
$$=30\bigl[30\cdot 2^{29}-29\cdot 2^{28}\bigr]=30\cdot 2^{28}(60-29)=30\cdot 31\cdot 2^{28}=465\cdot 2^{29}.$$
Hence $\alpha=465.$
Correct Answer: 465