There are 99 sequences to sum up.
The n-th sequence looks like this:
n(n+1) + (n+1)(n+2) + ... + 99(100).
We can solve this by summation:
\(\quad n(n+1 ) + (n+1)(n+2) + \cdots + 99(100)\\ =(1\cdot2 + 2\cdot 3 + ... + 99(100)) - (1\cdot 2 + 2\cdot 3 + ... + n(n-1))\\ =\displaystyle \sum^{99}_{k=1} (k^2 + k) - \sum^{n-1}_{k=1}(k^2 + k)\\ = \dfrac{99(100)(199)}{6} + \dfrac{99(100)}{2} - \dfrac{n(n-1)(2n-1)}{6} - \dfrac{n(n-1)}{2}\\ = -\dfrac{n^3}{3} + \dfrac{n}{3} + 333300\)
Now we need to find the sum of this expression from n = 1 to p, where p is some positive integer.
\(\displaystyle\sum^{p}_{n=1} \left(\dfrac{-n^3}{3} + \dfrac{n}{3} + 333300\right)\\ = \dfrac{p(-p^3-2p^2+p+3999602)}{12}\)
We sum this from p = 1 to p = 100 and we get 1,508,082,510.
:)
To Guest: The first sequence is (2 + 3 + 4 + ... + 100) + (6 + 8 + 10 + ... + 200) + ... + (98 * 99 + 98*100 + 99*100).
