Using mathematical induction, the numbers a_n are defined by a₀=1, a_ n+1 =3 n^2+n+a_n, (n 0) . Then, a_n is equal to

Step-by-step solution

The recurrence relation is given by a_ n+1 = a_n + 3n^2 + n with a₀ = 1 . We can write the general term a_n as a summation: a_n = a₀ + _ k=0 ^ n-1 (a_ k+1 - a_k) = 1 + _ k=0 ^ n-1 (3k^2 + k) . Using the standard summation formulas _ k=0 ^ n-1 k^2 = (n-1)n(2n-1) 6 and _ k=0 ^ n-1 k = (n-1)n 2 , we substitute these into the expression for a_n : a_n = 1 + 3 [ (n-1)n(2n-1) 6 ] + (n-1)n 2 . Simplifying the expression: a_n = 1 + n(n-1)(2n-1) 2 + n(n-1) 2 = 1 + n(n-1) 2 [ (2n-1) + 1 ] . a_n = 1 + n(n-1) 2 [ 2n ] = 1 + n^2