JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q22

MathematicsInteger+4 / -1Question 22 of 75

If $\displaystyle\sum_{k=1}^na_k=6n^3$ , then $\displaystyle\sum_{k=1}^6\left(\frac{a_{k+1}-a_k}{36}\right)^2$ is equal to ________.

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Answer and worked solution

Official answer: 91

Terms. $a_k=6k^3-6(k-1)^3=6(3k^2-3k+1)$ , so $a_{k+1}-a_k=6\cdot6k=36k$ .

Sum. $\displaystyle\sum_{k=1}^6\left(\frac{36k}{36}\right)^2=\sum_{k=1}^6k^2=91$ .

Answer: 91

JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 22