Question

The sum to n terms of the series \[\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+......\]

Answer 1

\sqrt{2} *(n*(n+1)/2)

Answer 2

Take root 2 common

Answer 3

So, it becomes root2(1+2+3+4+...)

Answer 4

lol! so simple, i can't believe you couldn't do this.....

Answer 5

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}