The modified problem of chess board [part-1]

here $100\times100$ squares appears 1 time. $99\times99$ squares appears 4 times $98\times98$ square appears 9 times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $1\times1$ appears 10000[100 X100] times therefore we must implicate $\sum n^2$ formula $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ $=>\sum n^2 = \frac{100(100+1)(200+1)}{6}$ $=>\sum n^2 = \frac{100\times101\times201}{6}$ $=>\sum n^2=338350$ therefore the no. of squares in a $100\times100$ chess board is 338350

By using the formula to find the sum of squares of first n natural numbers , we et the answer as 338350.

S (n) = (1/ 6) n (n + 1) (2 n + 1)

S (100) = 338350