1. ## Algebra....

I Really have no idea about how to approach this one...

2/[y(y+1)] can be expressed in form 2/y - 2/y+1

thanks

2. Originally Posted by Ant
I Really have no idea about how to approach this one...

2/[y(y+1)] can be expressed in form 2/y - 2/y+1

thanks
That's called "partial fractions". Suppose $\frac{2}{y(y+1)}= \frac{A}{y}+ \frac{B}{y+1}$. Multiply on both sides by y(y+1) to get rid of the fractions: $2= A(y+1)+ By$. That must be true for all y so you can take any value for y you want. In particular, taking y= 0 give 2= A and taking y= -1 gives 2= -B.