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Yahoo!
The sum to n terms of the series \[\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+......\]
\sqrt{2} *(n*(n+1)/2)
Take root 2 common
So, it becomes root2(1+2+3+4+...)
lol! so simple, i can't believe you couldn't do this.....
sqrt(2)+sqrt(2*4)+sqrt(2*9)+sqrt(2*16)+........ =sqrt(2)+2*sqrt(2)+3*sqrt(2)+4*sqrt(2)+...... =sqrt(2)*{1+2+3+4+....} =sqrt(2)*{n*(n+1)/2}