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

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- Oct 30th 2008, 06:41 AMAntAlgebra....
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 - Oct 30th 2008, 12:49 PMHallsofIvy
That's called "partial fractions". Suppose $\displaystyle \frac{2}{y(y+1)}= \frac{A}{y}+ \frac{B}{y+1}$. Multiply on both sides by y(y+1) to get rid of the fractions: $\displaystyle 2= A(y+1)+ By$. 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.