What is the sum of the squares of integers starting with 10 and till 48
Find the sum of the squares of n from n=10 to n=48:
The sum of the first n squares is `S_n=(n(n+1)(2n+1))/6` .
To find the sum from n=10 top n=48 we take `S_48-S_9`
** Take `S_9` because we want to include `10^2` ; we want to discard n=1-9 since we counted them in `S_48` , but we do not want them in the final sum.**
The sum is 37739.