If sum_{r=1}^{30} frac{r^2({}^{30}C_r)^2}{{}^ | vedclass

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(D) We have the expression $S = \sum_{r=1}^{30} \frac{r^2({}^{30}C_r)^2}{{}^{30}C_{r-1}}$.Using the property $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{

Vedclass · Accepted Answer

$465$

Vedclass · Answer

(D) We have the expression $S = \sum_{r=1}^{30} \frac{r^2({}^{30}C_r)^2}{{}^{30}C_{r-1}}$.Using the property $\frac{{}^{n}C_r}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{{}^{30}C_r}{{}^{30}C_{r-1}} = \frac{31-r}{r}$.Thus,the term becomes $r^2 \cdot {}^{30}C_r \cdot \frac{31-r}{r} = r(31-r) {}^{30}C_r$.Since $r \cdot {}^{30}C_r = 30 \cdot {}^{29}C_{r-1}$,the sum is $\sum_{r=1}^{30} 30 \cdot {}^{29}C_{r-1} (31-r)$.Let $k = r-1$,then $r = k+1$. As $r$ goes from $1$ to $30$,$k$ goes from $0$ to $29$.$S = 30 \sum_{k=0}^{29} {}^{29}C_k (30-k) = 30 \left( 30 \sum_{k=0}^{29} {}^{29}C_k - \sum_{k=0}^{29} k \cdot {}^{29}C_k \right)$.Using $\sum_{k=0}^{n} {}^{n}C_k = 2^n$ and $\sum_{k=0}^{n} k \cdot {}^{n}C_k = n \cdot 2^{n-1}$:$S = 30 \left( 30 \cdot 2^{29} - 29 \cdot 2^{28} \right) = 30 \cdot 2^{28} (60 - 29) = 30 \cdot 31 \cdot 2^{28} = 15 \cdot 31 \cdot 2^{29} = 465 \cdot 2^{29}$.Therefore,$\alpha = 465$.

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

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