Let a sequence be defined by a₁ = 4 and a_ k+1 - a_k = 6k + 2 for all k 1 . If S_n = _ k=1 ^n a_k and n is a positive integer such that 2900 < S_n < 3000 , then the value of n is ____________.

Step-by-step solution

Given a₁ = 4 and a_ k+1 - a_k = 6k + 2 . We can find the k -th term a_k using the method of differences: a_k = a₁ + _ j=1 ^ k-1 (a_ j+1 - a_j) a_k = 4 + _ j=1 ^ k-1 (6j + 2) a_k = 4 + 6 (k-1)k 2 + 2(k-1) a_k = 4 + 3k^2 - 3k + 2k - 2 a_k = 3k^2 - k + 2 Now, we compute the sum S_n = _ k=1 ^n a_k : S_n = _ k=1 ^n (3k^2 - k + 2) S_n = 3 n(n+1)(2n+1) 6 - n(n+1) 2 + 2n S_n = n(n+1)(2n+1) 2 - n(n+1) 2 + 2n S_n = n(n+1) 2 [2n+1 - 1] + 2n S_n = n^2(n+1) + 2n = n^3 + n^2 + 2n We are given that 2900 So, 2900 Let us test integ