### Instruction: What is the value of the sum $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{n(n+1)} +\cdots + \frac{1}{9900}$? Express your answer as a common fraction.### Response: I notice that the terms in the sum have a pattern: they are of the form $\frac{1}{n(n+1)}$ for some positive integer $n$.
I wonder if I can find a simpler way to write each term, maybe by using partial fraction decomposition.
I recall that partial fraction decomposition is a method to write a single fraction as a sum of simpler fractions with different denominators.
I try to apply this method to the general term $\frac{1}{n(n+1)}$.
I write it as $\frac{A}{n} + \frac{B}{n+1}$ for some constants $A$ and $B$.
I multiply both sides by $n(n+1)$ to get rid of the denominators.
I get $1 = A(n+1) + Bn$.
I want to find the values of $A$ and $B$ that make this equation true for any $n$.
I notice that if I plug in $n=0$, I get $1=A$, so $A=1$.
I also notice that if I plug in $n=-1$, I get $1=-B$, so $B=-1$.
Therefore, I can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
I substitute this expression into the original sum and see if I can simplify it.
I get $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{9900} = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + \cdots + (\frac{1}{99} - \frac{1}{100})$.
I notice that most of the terms cancel out, leaving only the first and the last terms.
I get $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{9900} = \frac{1}{2} - \frac{1}{100}$.
I find a common denominator and simplify the fraction.
I get $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \cdots + \frac{1}{9900} = \frac{50-1}{100} = \frac{49}{100}$.