Finding 'a' and 'b' (arithmetic)

Question

Finding 'a' and 'b': $$\frac{1}{n(n+1)} = \frac{a}{n}+\frac{b}{1+n}$$

That what I did: $$\frac{1}{n(n+1)} = \frac{a(1+n)+bn}{n(1+n)}$$ $$1 = a(1+n)+bn$$ $$1 = a+an+bn$$ $$1 = a+n(a+b)$$

Answer 1

your last line can be written as $$1=a+n(a+b)$$ comparing both sides we get $$a=1$$ and $a+b=0$ thus $b=-1$

Answer 2

Let me rephrase a bit:

For $n$, positive, integer, you want the equation to hold.

You find:

$1= a + n(a+b).$

You want this to hold for any n, positive, integer.

$ \rightarrow: $

1)$a+b=0.$ (No $n$-term on the LHS)

$ \rightarrow:$

2)$a =1.$