Question

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

Answer 1

Step 1: Find the general term $a_n$.
Let $S_n = \sum_{k=1}^{n} a_k = 6n^3$.
Then $a_n = S_n - S_{n-1} = 6n^3 - 6(n-1)^3$.
$a_n = 6 [n^3 - (n^3 - 3n^2 + 3n - 1)] = 6(3n^2 - 3n + 1)$.


Step 2: Calculate $a_{k+1} - a_k$.
$a_{k+1} = 6(3(k+1)^2 - 3(k+1) + 1) = 6(3k^2 + 6k + 3 - 3k - 3 + 1) = 6(3k^2 + 3k + 1)$.
$a_k = 6(3k^2 - 3k + 1)$.
$a_{k+1} - a_k = 6 [(3k^2 + 3k + 1) - (3k^2 - 3k + 1)] = 6(6k) = 36k$.


Step 3: Evaluate the final sum.
The expression is $\sum_{k=1}^{6} \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^{6} k^2$.
Using the formula for sum of squares: $\frac{n(n+1)(2n+1)}{6}$ for $n=6$:
$= \frac{6(7)(13)}{6} = 7 \times 13 = 91$
The answer is 91.