 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/10/a/solution

Change the summation index to

j = n + 1

. Then

j

runs from

- 99

100

. Each term

j^{3}

for

1 \leq j \leq 99

cancels its partner

(- j)^{3}

, and the zero term contributes nothing. The sole unpaired term is

10 0^{3}

, so

n = - 100 \sum 99 (n + 1)^{3} = 1 0^{6} .

 New to topics? Read the docs here!