If , then is equal to _______. [2026]

Question

If $\sum_{k = 1}^{n} a_{k} = 6 n^{3}$ , then $\sum_{k = 1}^{6} {(\frac{a_{k + 1} - a_{k}}{36})}^{2}$ is equal to _______. [2026]

Smart Achievers · Accepted Answer

(91)

Given, $\sum_{k = 1}^{n} a_{k} = 6 n^{3}$

Now, $a_{n} = \sum_{k = 1}^{n} a_{k} - \sum_{k = 1}^{n - 1} a_{k}, n \geq 2$

$= 6 n^{3} - 6 {(n - 1)}^{3}$

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

$\Rightarrow a_{n} = 6 (3 n^{2} - 3 n + 1), n \geq 2$

$∴ a_{k} = 6 (3 k^{2} - 3 k + 1)$

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

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

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

$= 6 [3 (2 k + 1) - 3] = 36 k$

$\Rightarrow \frac{a_{k + 1} - a_{k}}{36} = k$

Hence, $\sum_{k = 1}^{6} {(\frac{a_{k + 1} - a_{k}}{36})}^{2}$

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

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