Using mathematical induction, the numbers a_ n 's are defined by a₀=1, a_ n+1 =3 n²+n+a_ n , (n 0) . Then, a_ n is equal to

Step-by-step solution

Given the recurrence relation a_ n+1 = a_ n + 3n² + n with a₀ = 1 . We can write the terms as: a_ n = a₀ + _ k=0 ^ n-1 (a_ k+1 - a_ k ) a_ n = 1 + _ k=0 ^ n-1 (3k² + k) Using the standard summation formulas _ k=0 ^ n-1 k² = (n-1)n(2n-1) 6 and _ k=0 ^ n-1 k = (n-1)n 2 : a_ n = 1 + 3 [ (n-1)n(2n-1) 6 ] + (n-1)n 2 a_ n = 1 + n(n-1)(2n-1) 2 + n(n-1) 2 a_ n = 1 + n(n-1) 2 [ (2n-1) + 1 ] a_ n = 1 + n(n-1)(2n) 2 a_ n = 1 + n²(n-1) a_ n = 1 + n³ - n² Answer: n³-n²+1