If _r=1^30r^2(^30C_r)^2^30C_r-1= 2^29, then is equal to _________.

Question

If $\sum _{r=1}^{30}\frac{{r}^{2}{({}^{30}{C}_{r})}^{2}}{{}^{30}{C}_{r-1}}=\alpha 	imes {2}^{29}$, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 465

Accepted Answer

$\begin{gathered} \sum _{r=1}^{30}\frac{{r}^{2}{({}^{30}{C}_{r})}^{2}}{{}^{30}{C}_{r-1}} \ =\sum _{r=1}^{30}{r}^{2}(\frac{31-r}{r})\cdot \frac{30!}{r!(30-r)!} \ (\because \frac{{}^{30}{C}_{r}}{{}^{30}{C}_{r-1}}=\frac{30-r+1}{r}=\frac{31-r}{r}) \ =\sum _{r=1}^{30}\frac{(31-r)30!}{(r-1)!(30-r)!} \ =30\sum _{r=1}^{30}\frac{(31-r)29!}{(r-1)!(30-r)!} \ =30\sum _{r=1}^{30}(30-r+1{)}^{29}{C}_{30-r} \ =30(\sum _{r=1}^{30}(31-r{)}^{29}{C}_{30-r}+\sum _{r=1}^{30}{}^{29}{C}_{30-r}) \ =30(29	imes {2}^{28}+{2}^{29})=30(29+2){2}^{28} \ =15	imes 31	imes {2}^{29} \ =465({2}^{29}) \ \alpha =465 \end{gathered}$

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