If sum_{k=1}^{n} a_k = 6n^3, then sum_{k=1}^{ | vedclass

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(A) Given $S_n = \sum_{k=1}^{n} a_k = 6n^3$.We know that $a_n = S_n - S_{n-1}$ for $n > 1$.$a_n = 6n^3 - 6(n-1)^3 = 6(n^3 - (n^3 - 3n^2 + 3n - 1)) = 6

Vedclass · Accepted Answer

$91$

Vedclass · Answer

(A) Given $S_n = \sum_{k=1}^{n} a_k = 6n^3$.We know that $a_n = S_n - S_{n-1}$ for $n > 1$.$a_n = 6n^3 - 6(n-1)^3 = 6(n^3 - (n^3 - 3n^2 + 3n - 1)) = 6(3n^2 - 3n + 1) = 18n^2 - 18n + 6$.Now, calculate $a_{k+1} - a_k$:$a_{k+1} - a_k = [18(k+1)^2 - 18(k+1) + 6] - [18k^2 - 18k + 6]$$= 18(k^2 + 2k + 1 - k^2) - 18(k + 1 - k) = 18(2k + 1) - 18 = 36k$.Substitute this into the given expression:$\sum_{k=1}^{6} \left(\frac{36k}{36}\right)^2 = \sum_{k=1}^{6} k^2$.Using the formula for the sum of squares of the first $n$ natural numbers, $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.For $n=6$, $\sum_{k=1}^{6} k^2 = \frac{6(6+1)(2 \cdot 6 + 1)}{6} = 7 \cdot 13 = 91$.

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

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