Know your tricks (Part 3)

The triangle number sequence follows as $(1+2+...+n)$ for the $nth$ number. So, for the triangular numbers upto 1000, there will be 1000 1's, 999 2's .... 2 999's and 1 1000.

$\sum\limits_{i=1}^{1000} (1001-n)(n)$ $\sum\limits_{i=1}^{1000} (1001n) - \sum\limits_{i=1}^{1000}(n^2)$ $1001\times\frac{(1000)(1001)}{2} - \frac{1000(1001)(2001)}{2}$ $\boxed{167167000}$

The sum is equal to n/2(n + 1) summed from n = 1 to n =1000 = (1/2)[sum of n^2 + n} = (1/2)[(1000)(1001)(2001)/6 + (1000)(1001)/2]= 167,167,000. Ed Gray.