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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

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Do you know what partial fractions are? – Calvin Lin Aug 1 '13 at 15:10

Hint: $$\frac{2d}{a_na_{n+1}a_{n+2}} = \frac{1}{a_na_{n+1}} - \frac{1}{a_{n+1}a_{n+2}}$$

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I think something wrong here as we see : $\frac{1}{a_1a_2}-\frac{1}{a_2a_3} = \frac{a_2a_3 -a_1a_2}{a_1a_2aa_3} = \frac{a_2(a_3-a_1)}{a_1a_2a_3} \neq \frac{2d}{a_1a_2a_3}$ – sultan Aug 1 '13 at 16:03
@sultan $\frac{a_2a_3}{a_1a_2a_3}\neq \frac{1}{a_1a_2}$. Try again. – Thomas Andrews Aug 1 '13 at 16:11
I mean what you have written should give after solving L.H.S = R.H.S but it is not so...which is $\frac{2d}{a_1a_3} \neq L.H.S.$ – sultan Aug 1 '13 at 16:20
Can we write it this way : $\frac{a_2-a_1}{a_1a_2a_3}+\frac{a_3-a_2}{a_1a_2a_3} = \frac{1}{a_1a_2}-\frac{1}{a_2a_3}$.....what next then... – sultan Aug 1 '13 at 16:27
You said "we see..." and then stated an equation that wasn't true. My formula is accurate. It's not clear to me what your objection is. – Thomas Andrews Aug 1 '13 at 16:57

Hint: The partial fraction of $\frac{1}{ (a+d)(a+2d)(a+3d)}$ is

$$ \frac{1}{d^2}\left( \frac{1}{2(a+d)} - \frac{1}{(a+2d)} + \frac{1}{2(a+3d)} \right). $$

Now, apply the telescoping series. Observe that terms do cancel out since $\frac{1}{2} -1 + \frac{1}{2} = 0 $.

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Do we have some method shorter than partial fraction...thanks.... – sultan Aug 1 '13 at 16:10
After doing the partial fraction of this I am getting the following three equation in A,B,C which is : A+B+C=0 ; 5A+4B+3C =0 ; 6A+3B+2C=1 which gives $-\frac{1}{2} -1+\frac{1}{2}$ and not $\frac{1}{2} -1 +\frac{1}{2}$ please look into this... thanks.. – sultan Aug 2 '13 at 3:37

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