Let S_k = frac{1 + 2 + 3 + .... + k}{k}. If S | vedclass

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(C) Given $S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.We need to find $A$ such that $\sum_{k=1}^{10} S_k^2 = \frac{5}{12}A$.$\sum_{k=1}^{10} \left( \fra

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$303$

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(C) Given $S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2}$.We need to find $A$ such that $\sum_{k=1}^{10} S_k^2 = \frac{5}{12}A$.$\sum_{k=1}^{10} \left( \frac{k+1}{2} ight)^2 = \frac{5}{12}A$$\frac{1}{4} \sum_{k=1}^{10} (k+1)^2 = \frac{5}{12}A$$\frac{1}{4} (2^2 + 3^2 + .... + 11^2) = \frac{5}{12}A$We know $\sum_{n=1}^{n} n^2 = \frac{n(n+1)(2n+1)}{6}$.So,$\sum_{k=1}^{11} k^2 = \frac{11(12)(23)}{6} = 11 	imes 2 	imes 23 = 506$.Therefore,$2^2 + 3^2 + .... + 11^2 = 506 - 1^2 = 505$.Substituting this back: $\frac{1}{4} (505) = \frac{5}{12}A$.$A = 505 	imes \frac{12}{4 	imes 5} = 505 	imes \frac{3}{5} = 101 	imes 3 = 303$.

Let $S_k = \frac{1 + 2 + 3 + .... + k}{k}$. If $S_1^2 + S_2^2 + ....... + S_{10}^2 = \frac{5}{12}A$,then $A$ is equal to:

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