The value of lim n → ∞ ∑ k = 1 n n 3 n 2 + k 2 n 2 + 3 k 2 is :

Step-by-step solution

Given, lim n → ∞ ∑ k = 1 n n 3 n 2 + k 2 n 2 + 3 k 2 = lim n → ∞ 1 n ∑ k = 1 n 1 1 + k 2 n 2 1 + 3 · k 2 n 2 Now, using limit as a sum integral we get, = ∫ 0 1 d x 1 + x 2 1 + 3 x 2 = 1 2 ∫ 0 1 3 1 + 3 x 2 - 1 1 + x 2 d x = 1 2 ∫ 0 1 1 1 3 2 + x 2 - 1 1 + x 2 d x = 1 2 3 tan - 1 3 x - tan - 1 x 0 1 = 1 2 3 tan - 1 3 - tan - 1 1 = 1 2 3 · π 3 - π 4 = 1 2 π 3 - π 4 = 13 π 8 · ( 4 3 + 3 )