Summation of 3n + 2 from n = 1 to n = 14 = (3(1) + 2) + (3(2) + 2) + (3(3) + 2) + . . . + (3(14) + 2) = 5 + 8 + 11 + ... + 44 ia an arithmetic progression with first term (a) = 5, common difference (d) = 3 and last term (l) = 44 and n = 14
Sn = n/2(a + l) = 14/2(5 + 44) = 7(49) = 343
Therefore, the required summation is 343.
