_ n 1 n^3+1 + 4 n^3+1 + 9 n^3+1 + + n^2 n^3+1 =

Step-by-step solution

To evaluate the limit, we express the sum using summation notation: _ n 1 n^3+1 (1^2 + 2^2 + 3^2 + + n^2) . The sum of the first n squares is _ k=1 ^ n k^2 = n(n+1)(2n+1) 6 , which gives: _ n 1 n^3+1 n(n+1)(2n+1) 6 = _ n 2n^3 + 3n^2 + n 6n^3 + 6 . Dividing numerator and denominator by n^3 : _ n 2 + 3 n + 1 n^2 6 + 6 n^3 = 2 + 0 + 0 6 + 0 = 1 3 . The limit is 1 3 , so the answer is B .