Let binom{n}{k} = frac{n!}{k!(n-k)!}. Then th | vedclass

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(B) We know that $\sum_{k=0}^{n} \binom{n}{k} k^2 = n(n+1) 2^{n-2}$.For $n=10$,the sum is $\sum_{k=0}^{10} \binom{10}{k} k^2 = 10(11) 2^{10-2} = 110 \

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$(2.5, 2.6)$

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(B) We know that $\sum_{k=0}^{n} \binom{n}{k} k^2 = n(n+1) 2^{n-2}$.For $n=10$,the sum is $\sum_{k=0}^{10} \binom{10}{k} k^2 = 10(11) 2^{10-2} = 110 	imes 2^8$.Now,we calculate the given expression:$\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2 = \frac{110 	imes 2^8}{2^{10}} = \frac{110}{2^2} = \frac{110}{4} = 27.5$.Since the question asks for the interval,and $27.5$ is not in the original options provided,we note that the calculation yields $27.5$,which lies in the interval $(27, 28)$.

Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

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