If _ k=1 ^ n a_k = 6n^3 , then _ k=1 ⁶ ( a_ k+1 - a_k 36 )^2 is equal to _______.

Step-by-step solution

Given _ k=1 ^ n a_k = 6n^3 . Let S_n = 6n^3 . The n -th term of the sequence is given by a_n = S_n - S_ n-1 . a_n = 6n^3 - 6(n-1)^3 a_n = 6[n^3 - (n^3 - 3n^2 + 3n - 1)] a_n = 6(3n^2 - 3n + 1) Replacing n with k and k+1 , we get: a_k = 6(3k^2 - 3k + 1) a_ k+1 = 6(3(k+1)^2 - 3(k+1) + 1) = 6(3k^2 + 3k + 1) Now, find the difference a_ k+1 - a_k : a_ k+1 - a_k = 6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1) a_ k+1 - a_k = 6(6k) = 36k Substitute this into the given summation: _ k=1 ⁶ ( a_ k+1 - a_k 36 )^2 = _ k=1 ⁶ ( 36k 36 )^2 _