If $\sum_{r=1}^{30} \frac{r^2 (^{30}C_r)^2}{^{30}C_{r-1}} = \alpha \times 2^{29}$, then $\alpha$ is equal to
Solution:
Let the general term be $T_r = \frac{r^2 (^{30}C_r)^2}{^{30}C_{r-1}}$.
Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$, with $n=30$:
$$ T_r = r^2 \cdot ^{30}C_r \cdot \left( \frac{30-r+1}{r} \right) = r(31-r)^{30}C_r $$
We need to find sum $S = \sum_{r=1}^{30} r(31-r)^{30}C_r$.
Using property $r \cdot ^{n}C_r = n \cdot ^{n-1}C_{r-1}$:
$$ S = \sum_{r=1}^{30} (31-r) \left( 30 \cdot ^{29}C_{r-1} \right) $$
$$ S = 30 \sum_{r=1}^{30} (31-r) ^{29}C_{r-1} $$
Let $k = r-1$. When $r=1, k=0$ and when $r=30, k=29$.
$$ S = 30 \sum_{k=0}^{29} (30-k) ^{29}C_k $$
$$ S = 30 \left[ 30 \sum_{k=0}^{29} {^{29}C_k} – \sum_{k=0}^{29} k \cdot {^{29}C_k} \right] $$
Using standard summation results $\sum {^nC_k} = 2^n$ and $\sum k \cdot {^nC_k} = n \cdot 2^{n-1}$:
$$ S = 30 \left[ 30 \cdot 2^{29} – 29 \cdot 2^{28} \right] $$
$$ S = 30 \cdot 2^{28} \left[ 30(2) – 29 \right] $$
$$ S = 30 \cdot 2^{28} [60 – 29] = 30 \cdot 31 \cdot 2^{28} $$
$$ S = 15 \cdot 31 \cdot 2^{29} = 465 \cdot 2^{29} $$
Given $S = \alpha \times 2^{29}$, we get $\alpha = 465$.
Ans. (465)