Whole Number Problem

To put it in summation notation:
$\large\sum_{i=1}^n i = \dfrac{n*(n+1)}{2} \\ We~ have~ n=100 , ~so~ it~is =5050\\If~we~had~sum~of~squares ~of~1~to~n\\ \large\sum_{i=1}^n i^2 = \dfrac{n*(n+1)(2n+1)}{6} \\Sum ~of ~cubes~will~be~\large\sum_{i=1}^n i^3 =( \dfrac{n*(n+1)}{2})^2$

Pair the beginning and end terms (first & last, second & second-last. . . ).

Each of fifty such pairs adds to 101, or ‘100 + 1’ (1 + 100, 2 + 99. . . ).

Sum = 5050 (50 hundreds and 50 ones).

S = 1 + 2 + 3 + ... + 98 + 99 + 100

S = 100 + 99 + 98 + ... + 3 + 2+ 1

2 S = 100 (101)

S = 50 x 101 = 5050