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

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- Jan 13th 2010, 03:34 AMdifferentiatesimple substitution in formula
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 - Jan 13th 2010, 03:45 AMArchie Meade
$\displaystyle 2^2+4^2+6^2+........+100^2=[1(2)]^2+[2(2)]^2+[3(2)]^2+......+[50(2)]^2$

$\displaystyle =[1^2+2^2+3^2+.......+50^2]2^2$

The number of terms is 50. - Jan 13th 2010, 03:52 AMmathaddict
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 .