JEE Main 2026 Apr 2 Shift 1, Mathematics Q22: Summation of Series

JEE Main2026Apr 2, Shift 1Mathematics

If $k = 1 \sum n a_{k} = 6 n^{3}$ , then $k = 1 \sum 6 (\frac{a _{k + 1} - a _{k}}{36})^{2}$ is equal to ____.

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Solution

From the partial sum formula, $a_{n} = S_{n} - S_{n - 1} = 6 n^{3} - 6 (n - 1)^{3}$ for $n \geq 2$ (and $a_{1} = 6$ which also fits). Expanding:

$a_{n} = 6 [n^{3} - (n - 1)^{3}] = 6 [3 n^{2} - 3 n + 1]$

Then $a_{k + 1} - a_{k} = 6 [3 (k + 1)^{2} - 3 (k + 1) + 1] - 6 [3 k^{2} - 3 k + 1] = 6 \cdot 6 k = 36 k$ .

Hence $(\frac{a _{k + 1} - a _{k}}{36})^{2} = k^{2}$ , and

$\sum_{k = 1}^{6} k^{2} = \frac{6 \cdot 7 \cdot 13}{6} = 91$ .

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