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: Understanding the Concept:
The sum of the first \( n \) terms is given as \( S_n = 6n^3 \).
We need to find the general term \( a_n \) using \( a_n = S_n - S_{n-1} \).
Step 2: Key Formula or Approach:
\( a_n = S_n - S_{n-1} = 6n^3 - 6(n-1)^3 \).
Step 3: Detailed Explanation:
\( a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)] = 6(3n^2 - 3n + 1) \).
Now find \( a_{k+1} - a_k \):
\( a_{k+1} = 6[3(k+1)^2 - 3(k+1) + 1] = 6(3k^2 + 6k + 3 - 3k - 3 + 1) = 6(3k^2 + 3k + 1) \).
\( a_{k+1} - a_k = 6[ (3k^2 + 3k + 1) - (3k^2 - 3k + 1) ] = 6(6k) = 36k \).
The required summation is:
\( \sum_{k=1}^6 \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^6 k^2 \).
Sum of squares of first 6 natural numbers:
\( \sum k^2 = \frac{n(n+1)(2n+1)}{6} = \frac{6(7)(13)}{6} = 91 \).
Step 4: Final Answer:
The value of the expression is 91.