Here's the question you clicked on:
TaylorBaugher2
What is the sum of all square numbers from 0 to 101?
n(n+1)(2n+1)/6 put n =101.
sum of squares of first 'n' natural or whole nos. is \[\sum_{i=1}^{n} i^2 = n(n+1)(2n+1)/6\] so plug in n=101, calculate and enjoy!!
n(n+1)(2n+1)/6 put n =101.
what you wrote - that's for the first 'n' nos. @KIT3 , not the sum of *square of each no.*