JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

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Explanation

\therefore S_{k} = \frac{1+2+\dots+k}{k}

= \frac{k(k+1)}{2k} = \frac{k+1}{2}

\Rightarrow S_{k}^{2} = \left(\frac{k+1}{2}\right)^{2} = \frac{k^{2}+1+2k}{4}

\Rightarrow \sum_{j=1}^{n} S_{j}^{2} = \frac{1}{4}\left[\sum_{j=1}^{n} k^{2}+\sum_{j=1}^{n} 1+2 \sum_{j=1}^{n} k\right]

= \frac{1}{4}\left[\frac{n(n+1)(2n+1)}{6}+n+\frac{2n(n+1)}{2}\right]

= \frac{n}{4}\left[\frac{(n+1)(2n+1)}{6}+1+n+1\right]

= \frac{n}{24}\left[2n^{2}+3n+1+6+6n+6\right]

= \frac{n}{24}\left[2n^{2}+9n+13\right]

= \frac{n}{24}\left[2n^{2}+9n+13\right]

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