Answer
The simplified solution is $\frac{1}{x}-\frac{1}{x+1}$ and its sum is $\frac{99}{100}$.
Work Step by Step
Let us consider the rational expression.
$\begin{align}
& \frac{1}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1} \\
& =\frac{A\left( x+1 \right)+Bx}{x\left( x+1 \right)}
\end{align}$
$1=A\left( x+1 \right)+Bx$ (I)
Substitute the value of $x=-1$ in the equation (I),
$\begin{align}
& 1=B\left( -1 \right) \\
& B=-1
\end{align}$
Again, putting the value of $x=0$ in the equation (I),
$\begin{align}
& 1=A\left( 1 \right) \\
& A=1
\end{align}$
Therefore, $\frac{1}{x\left( x+1 \right)}=\frac{1}{x}-\frac{1}{x+1}$ is the simplified form of the result
By using the above result to find the sum of the series,
$\begin{align}
& \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{99\cdot 100} \\
& =\left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\cdots +\left( \frac{1}{99}-\frac{1}{100} \right) \\
& =\frac{1}{1}-\frac{1}{100} \\
& =\frac{99}{100}
\end{align}$
Thus, the sum is $\frac{99}{100}$.
