If sum_{k=1}^{30} k left({ }^{30} C _kright)^ | vedclass

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(C) Let $S = \sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2$. Using the property ${ }^{n} C _k = { }^{n} C _{n-k}$,we have ${ }^{30} C _k = { }^{30} C

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let $S = \sum_{k=1}^{30} k \left({ }^{30} C _kight)^2$. Using the property ${ }^{n} C _k = { }^{n} C _{n-k}$,we have ${ }^{30} C _k = { }^{30} C _{30-k}$. Thus,$S = \sum_{k=1}^{30} k \left({ }^{30} C _{30-k}ight)^2$. Let $j = 30-k$,then $k = 30-j$. As $k$ goes from $1$ to $30$,$j$ goes from $29$ to $0$. $S = \sum_{j=0}^{29} (30-j) \left({ }^{30} C _jight)^2 = 30 \sum_{j=0}^{29} \left({ }^{30} C _jight)^2 - \sum_{j=0}^{29} j \left({ }^{30} C _jight)^2$. Since $30 \left({ }^{30} C _{30}ight)^2 = 30$,we can write $S = 30 \sum_{j=0}^{30} \left({ }^{30} C _jight)^2 - S - 30 \left({ }^{30} C _{30}ight)^2 + 30 \left({ }^{30} C _{30}ight)^2$ is not needed,simply: $2S = 30 \sum_{k=0}^{30} \left({ }^{30} C _kight)^2 - 30 \left({ }^{30} C _{30}ight)^2 + 30 \left({ }^{30} C _{30}ight)^2$ is incorrect. Correct approach: $S = \sum_{k=1}^{30} k \left({ }^{30} C _kight) \left({ }^{30} C _kight) = \sum_{k=1}^{30} k \left({ }^{30} C _kight) \left({ }^{30} C _{30-k}ight) = 30 \sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k}$. Using Vandermonde's Identity,$\sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k} = { }^{59} C _{29}$. $S = 30 	imes { }^{59} C _{29} = 30 	imes \frac{59!}{29! 30!} = 30 	imes \frac{59! 	imes 30}{30! 30!} = 15 	imes \frac{60!}{30! 30!}$. Thus,$\alpha = 15$.

If $\sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2 = \frac{\alpha 60 !}{(30 !)^2}$,then $\alpha$ is equal to

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