If sum_{k=1}^{10} k^{2} binom{10}{k}^{2} = 22 | vedclass

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(B) We are given the sum $S = \sum_{k=1}^{10} k^{2} \binom{10}{k}^{2}$.Using the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$,we have $k \binom{10}{

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$221$

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(B) We are given the sum $S = \sum_{k=1}^{10} k^{2} \binom{10}{k}^{2}$.Using the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$,we have $k \binom{10}{k} = 10 \binom{9}{k-1}$.Substituting this into the sum:$S = \sum_{k=1}^{10} (10 \binom{9}{k-1})^{2} = 100 \sum_{k=1}^{10} \binom{9}{k-1}^{2}$.Let $j = k-1$,then as $k$ goes from $1$ to $10$,$j$ goes from $0$ to $9$:$S = 100 \sum_{j=0}^{9} \binom{9}{j}^{2}$.Using the identity $\sum_{j=0}^{n} \binom{n}{j}^{2} = \binom{2n}{n}$,we get:$S = 100 \binom{18}{9} = 100 	imes 48620 = 4862000$.Given $S = 22000 L$,we have $22000 L = 4862000$.$L = \frac{4862000}{22000} = \frac{4862}{22} = 221$.

If $\sum_{k=1}^{10} k^{2} \binom{10}{k}^{2} = 22000 L$,then $L$ is equal to $.....$

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