Evaluate the summation sum_n=120 (1)/(n(n+1)) using partial fraction decomposition.

We can write $\frac{1}{n(n+1)}$ as $\frac{A}{n} + \frac{B}{n+1}$.
Multiplying by $n(n+1)$ gives $1 = A(n+1) + Bn$.
Setting $n=0$, we get $1 = A(1) \Rightarrow A=1$.
Setting $n=-1$, we get $1 = B(-1) \Rightarrow B=-1$.
So, the expression becomes:

The summation is $\sum_{n=1}^{20} \left(\frac{1}{n} - \frac{1}{n+1}\right)$.
Let's list the first few terms and the last term:
For $n=1$: $\left(1 - \frac{1}{2}\right)$
For $n=2$: $\left(\frac{1}{2} - \frac{1}{3}\right)$
For $n=3$: $\left(\frac{1}{3} - \frac{1}{4}\right)$
...
For $n=20$: $\left(\frac{1}{20} - \frac{1}{21}\right)$

This is a telescoping series, meaning most terms will cancel out:

The $-\frac{1}{2}$ cancels with $+\frac{1}{2}$, $-\frac{1}{3}$ cancels with $+\frac{1}{3}$, and so on.
The sum simplifies to the first term of the first parenthesis and the last term of the last parenthesis:

The final answer is $\text{D) } \frac{20}{21}$.