The sum of the following series 1 + 6 + 9 1 2 + 2 2 + 3 2 7 + 12 1 2 + 2 2 + 3 2 + 4 2 9 + 15 ( 1 2 + 2 2 + … + 5 2 ) 11 + . . . . up to 15 terms, is:

Step-by-step solution

t n = 3 n 1 2 + 2 2 + 3 2 + … + n 2 2 n + 1 = 3 n × n n + 1 2 n + 1 6   2 n + 1 = 1 2   n 3 + n 2 ∴ S 15 = ∑ n = 1 15 t n = ∑ n = 1 15 1 2 n 3 + n 2 = 1 2 ∑ n = 1 15 n 3 + ∑ n = 1 15 n 2 = 1 2 × 15 × 16 2 2 + 15 × 16 × 31 6   ∑ n 3 = n n + 1 2 2 ,   ∑ n 2 = n n + 1 n + 2 6 = 7200 + 620 = 7820