given that:
$\displaystyle 1^2 + 2^2 + 3^2 +.... + n^2 = \frac{n(n+1)(2n+1)}{6}$
then evaluate $\displaystyle 2^2 + 4^2 + 6^2 + ...... + 100^2 $
looks easy... but i can't get it
HI
another approach , of course Archie's one is better ,
$\displaystyle 1^2 + 2^2 + 3^2 +.... + n^2 = \frac{n(n+1)(2n+1)}{6}$ , Put n=100 , the sum of this would be 338350
then $\displaystyle 1^2+3^2+5^2+...+(2n-1)^2=\sum^{n}_{k=1}(2k-1)^2$
$\displaystyle \sum^{n}_{k=1}(4k^2-4k+1)=\frac{2n(n+1)(2n+1)}{3}-2n(n+1)+n$
put n=50 , and this sum is 166650
take their difference then .