# Prove that $\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$

There is this question in one of my math textbooks which I can't seem to figure out how to solve, it'd be awesome if you could help me :

If $a_1,a_2,a_3,...,a_n$ IS an arithmetic progression and $a_n$ is NOT equal to 0 then prove the following statement :

$\frac {1}{a_1a_2} + \frac {1}{a_2a_3} + \frac {1}{a_3a_4} + ... + \frac {1}{a_{n-1}a_n} = \frac {n-1}{a_1a_n}$

• You mean that $a_1, a_2, \ldots, a_n$ are an arithmetic progression? You should also require that none of the $a_i$ is zero. Oct 3, 2014 at 10:32
• oops yea, my bad (fixed it) Oct 3, 2014 at 10:46
• The tag (theorem-provers) is for questions about software designed for checking formal proofs or assisting with writing them, see the tag-wiki. It is not intended for all questions which are about proofs of theorems. Feb 9, 2015 at 14:45

Here's sketch, not complete proof.

Since $A$ is an arithmetic progression, Let $a_{i+1}-a_i = d$

\begin{align}\dfrac{1}{a_ia_{i+1}} &= \dfrac{d}{d}\dfrac{1}{a_ia_{i+1}} \\&= \dfrac{1}{d}\dfrac{a_{i+1}-a_i}{a_ia_{i+1}} \qquad \text{since } a_{i+1}-a_i = d \\&= \dfrac{1}{d}\left(\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}}\right)\end{align}

Therefore, \begin{align} \dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+\cdots + \dfrac{1}{a_{n-1}a_n}&= \sum\limits_{i = 1}^{n-1}\dfrac{1}{a_ia_{i+1}} \\&= \dfrac{1}{d}\sum\limits_{i = 1}^{n-1}\dfrac{1}{a_i}- \dfrac{1}{a_{i+1}} \\&= \dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac{1}{a_2}+\dfrac{1}{a_2}-\cdots - \dfrac{1}{a_{n-1}} + \dfrac{1}{a_{n-1}}-\dfrac{1}{a_n} \right)\\&=\dfrac{1}{d}\left(\dfrac{1}{a_1}-\dfrac1{a_n}\right)\end{align}

Now, take LCM and use the fact that $a_n = a_1 +(n-1)d$

• That partially makes sense to me but I'd really appreciate it if you could provide a more detailed explanation. Oct 3, 2014 at 10:50
• @Ashkan Which step was not clear to you? Oct 3, 2014 at 10:50
• Never mind, I read it again and it all made perfect sense, Thank you very much for your answer Oct 3, 2014 at 10:58
• beautiful answer, Thanks again Oct 3, 2014 at 11:05
• @Ashkan You're welcome. Oct 3, 2014 at 11:06