Question

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 _______.

Answer 1

Correct Answer: 91

Step 1: Understanding the Concept:
We are given the sum of the first \(n\) terms \(S_n\) of a sequence. The general term \(a_n\) is found using \(a_n = S_n - S_{n-1}\). After finding \(a_n\), we compute the difference of consecutive terms and evaluate the given sum.
Step 2: Key Formula or Approach:
1. \(a_n = S_n - S_{n-1}\).
2. Standard sum: \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\).
Step 3: Detailed Explanation:
Given \(S_n = 6n^3\).
The general term \(a_n\) is:
\[ a_n = 6n^3 - 6(n-1)^3 = 6[n^3 - (n^3 - 3n^2 + 3n - 1)] \]
\[ a_n = 6(3n^2 - 3n + 1) = 18n^2 - 18n + 6 \].
Now, calculate the difference \(a_{k+1} - a_k\):
\[ a_{k+1} - a_k = [18(k+1)^2 - 18(k+1) + 6] - [18k^2 - 18k + 6] \]
\[ = 18((k+1)^2 - k^2) - 18((k+1) - k) \]
\[ = 18(2k+1) - 18 = 36k + 18 - 18 = 36k \].
Substitute this into the required summation:
\[ \sum_{k=1}^6 \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^6 k^2 \]
Using the sum of squares formula for \(n=6\):
\[ \sum_{k=1}^6 k^2 = \frac{6(6+1)(2 \times 6 + 1)}{6} = 7 \times 13 = 91 \].
Step 4: Final Answer:
The final sum is 91.