algebra precalculus - How to find the sum of the series by treating deonominator so that to split fraction $frac{1}{a_1a_2a_3} + frac{1}{a_2a_3a_4}+$......

This is a series in A.P ( Arithmetic Progression )

$\frac{1}{a_1a_2} + \frac{1}{a_2a_3}+\frac{1}{a_3a_4}+.......\frac{1} { a_{n}a_{n+1}}$ ( where $a_1 ,a_2,a_3.....$ are terms in A.P.)

When we do sum of such series then we use the common difference concept to split the denominator in two parts so that one part is negative and other is positive i.e.

$\frac{1}{d}[ \frac{a_2-a_1}{a_1a_2}+\frac{a_3-a_2}{a_2a_3}+........\frac{a_{n}-a_{n+1}}{a_{n}a_{n+1}}$ ] ( where d is common difference which is further equal to the difference of two consecutive terms)

=$\frac{1}{d}[ \frac{1}{a_1} -\frac{1}{a_2}+\frac{1}{a_2}-\frac{1}{a_3}+........\frac{1}{a_{n}} -\frac{1}{a_{n+1}}]$ which after adding the series give us

$\frac{1}{d}[ \frac{1}{a_1}+\frac{1}{a_{n+1}}]$ = $\frac{1}{d}[ \frac{a_{n+1}+a_1}{a_1a_{n+1}}]$ ( here $a_{n+1}$ is the (n+1)th term ]

My question is do we have some method with the help of which we can split the following series ( which is in Arithmetic progression) ...

$\frac{1}{a_1a_2a_3} + \frac{1}{a_2a_3a_4}+\frac{1}{a_3a_4a_5}+.......\frac{1}{a_{n}a_{n+1}a_{n+2}}$

If yes....then please suggest that method... thanks