If sum_{ r =1}^{30} frac{ r ^2left({ }^{30} C | vedclass

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Question

$\sum_{ r =1}^{30} \frac{ r ^2\left({ }^{30} C _{ r }\right)^2}{{ }^{30} C _{ r -1}}$
$=\sum_{ r =1}^{30} r ^2\left(\frac{31- r }{ r }\right) \cdot \

Vedclass · Accepted Answer

$465$

Vedclass · Answer

$\sum_{ r =1}^{30} \frac{ r ^2\left({ }^{30} C _{ r }ight)^2}{{ }^{30} C _{ r -1}}$
$=\sum_{ r =1}^{30} r ^2\left(\frac{31- r }{ r }ight) \cdot \frac{30!}{ r !(30- r )!}$
$\left(\because \frac{{ }^{30} C _{ r }}{{ }^{30} C _{ r -1}}=\frac{30- r +1}{ r }=\frac{31- r }{ r }ight)$
$=\sum_{ r =1}^{30} \frac{(31- r ) 30!}{( r -1)!(30- r )!}$
$=30 \sum_{ r =1}^{30} \frac{(31- r ) 29!}{( r -1)!(30- r )!}$
$=30 \sum_{ r =1}^{30}(30- r +1)^{29} C _{30- r }$
$=30\left(\sum_{ r =1}^{30}(31- r )^{29} C _{30- r }+\sum_{ r =1}^{30}{ }^{29} C _{30- r }ight)$
$=30\left(29 	imes 2^{28}+2^{29}ight)=30(29+2) 2^{28}$
$=15 	imes 31 	imes 2^{29}$
$=465\left(2^{29}ight)$
$\alpha=465$

If $\sum_{ r =1}^{30} \frac{ r ^2\left({ }^{30} C _{ r }\right)^2}{{ }^{30} C _{ r -1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to

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