Partial Fractions: Why does this shortcut method work? Suppose I want to resolve $1/{(n(n+1))}$ into a sum of partial fractions. I solve this by letting $1/{(n(n+1))} = {a/n} + {b/(n+1)}$ and then solving for $a$ and $b$, which in this case gives $a=1$ and $b=-1$.
But I learnt about a shortcut method. It says suppose $1/{(n(n+1))} = {a/n} + {b/(n+1)}$, then find $a$ by finding the value which makes its denominator in the RHS equal to $0$ and computing the LHS with the $0$ term (or $a$'s denominator in RHS) removed so we get $a = {1/(0+1)} = 1$ [as $n=0$], and we get $b = {1/(-1)} = -1$ [as $n+1=0$].
Another example, if I am not clear, suppose $$\frac{1}{n(n+1)(n+2)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2};$$ then
$$
\begin{eqnarray}
a &=& \frac{1}{(0+1)(0+2)}=\frac{1}{2}, \\
b &=& \frac{1}{(-1)(-1+2)}=-1, \\
c &=& \frac{1}{(-2)(-2+1)}=\frac{1}{2}.
\end{eqnarray}
$$
Why does this shortcut method work?

I'll just start with $a$ as an example. Multiply both sides by $n$ to get
$$\frac 1{(n+1)(n+2)} = a + \frac {bn}{n+1} + \frac {cn}{n+2}$$
Since this is true for all $n$, plug in $0$. The $b$ and $c$ terms become $0$, leaving $a$ in the formula described above:
$$\frac 1{(0+1)(0+2)} = a$$
