Let SK=1+2+……+KK and ∑j=1n Sj2=nABn2+Cn+D, where A,B,C,D∈ℕ and A has least value. Then

Sk=k+12;Sk2=14k2+2k+1⇒∑Sk2=14k(k+1)(2k+1)6+k(k+1)+k=k242k2+9k+13

Let SK=1+2+……+KK and ∑j=1n Sj2=nABn2+Cn+D, where A,B,C,D∈ℕ and A has least value. Then

Sk=k+12;Sk2=14k2+2k+1⇒∑Sk2=14k(k+1)(2k+1)6+k(k+1)+k=k242k2+9k+13

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$Let S_{K} = \frac{1 + 2 + \dots \dots + K}{K} and \sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} ({Bn}^{2} + Cn + D), where A, B, C, D \in ℕ and A has least value. Then$

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$A + B is divisible by D$

$A + B = 5 (D - C)$

$A + C + D is not divisible by B$

$A + B + C + D is divisible by 5$

answer is A.

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$\begin{array}{l} S_{k} = \frac{k + 1}{2}; S_{k}^{2} = \frac{1}{4} (k^{2} + 2 k + 1) \\ \Rightarrow \sum S_{k}^{2} = \frac{1}{4} [\frac{k (k + 1) (2 k + 1)}{6} + k (k + 1) + k] \\ = \frac{k}{24} [2 k^{2} + 9 k + 13] \end{array}$

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Let SK=1+2+……+KK and ∑j=1n Sj2=nABn2+Cn+D, where A,B,C,D∈ℕ and A has least value. Then

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$Let S_{K} = \frac{1 + 2 + \dots \dots + K}{K} and \sum_{j = 1}^{n} S_{j}^{2} = \frac{n}{A} ({Bn}^{2} + Cn + D), where A, B, C, D \in ℕ and A has least value. Then$