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

Concept: If the partial sum of a sequence is given as \[ S_n=\sum_{k=1}^{n} a_k \] then \[ a_n=S_n-S_{n-1} \]
Step 1:Find \(a_n\).} \[ S_n=6n^3 \] \[ a_n=S_n-S_{n-1} \] \[ =6n^3-6(n-1)^3 \] \[ =6[n^3-(n^3-3n^2+3n-1)] \] \[ =6(3n^2-3n+1) \] \[ =18n^2-18n+6 \]
Step 2:Compute \(a_{k+1}-a_k\).} \[ a_{k+1}=18(k+1)^2-18(k+1)+6 \] \[ =18k^2+18k+6 \] Thus \[ a_{k+1}-a_k \] \[ =(18k^2+18k+6)-(18k^2-18k+6) \] \[ =36k \]
Step 3:Substitute into the summation} \[ \frac{a_{k+1}-a_k}{36}=k \] Thus \[ \sum_{k=1}^{6}\left(\frac{a_{k+1}-a_k}{36}\right)^2 = \sum_{k=1}^{6} k^2 \] \[ =1^2+2^2+3^2+4^2+5^2+6^2 \] \[ =91 \] \[ \boxed{91} \]