Answers for "1^2+2^2+3^2+...+n^2"

1^2+2^2+3^2+...+n^2

1^2+2^2+3^2+...+n^2 = n (n+1) (2n+1)
					 _________________
                            6

Posted by: Guest on April-03-2022

1^3+2^3+3^3+...+n^3

1^3+2^3+3^3+...+n^3 = [n(n+1)/2]^2

Posted by: Guest on April-03-2022

Code answers related to "1^2+2^2+3^2+...+n^2"