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- Feb 14, 2012

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I was wondering if this is a very hard problem to solve algebraically, rather than by mere observation.

Problem:

Given

\(\displaystyle \frac{a_1}{2}+\frac{a_2}{3}+...+\frac{a_{2013}}{2014}=\frac{4}{3}\)

\(\displaystyle \frac{a_1}{3}+\frac{a_2}{4}+...+\frac{a_{2013}}{2015}=\frac{4}{5}\)

\(\displaystyle \frac{a_1}{4}+\frac{a_2}{5}+...+\frac{a_{2013}}{2016}=\frac{4}{7}\)

...

\(\displaystyle \frac{a_1}{2014}+\frac{a_2}{2015}+...+\frac{a_{2013}}{4026}=\frac{4}{4027}\)

Find

\(\displaystyle \frac{a_1}{3}+\frac{a_2}{5}+...+\frac{a_{2013}}{4027}\)

Attempt:

I first started out by using only two terms, $a_1$ and $a_2$ to find the values for each terms and then add one-third to the variable $a_1$ and one-fifth to the variable $a_2$ and I see that

\(\displaystyle \frac{a_1}{3}+\frac{a_2}{5}=\frac{24}{25}=1-\frac{1}{5^2}\)

I then played with three terms, $a_1, a_2$ and $a_3$ and get

\(\displaystyle \frac{a_1}{3}+\frac{a_2}{5}+\frac{a_3}{7}=\frac{48}{49}=1-\frac{1}{7^2}\)

and Et voila, we can conclude that \(\displaystyle \frac{a_1}{3}+\frac{a_2}{5}+...+\frac{a_{2013}}{4027}=1-\frac{1}{4027^2}\)

But I know this is a piece of sloppy working...could someone show me the "real" method to solve this kind of problem? Thanks.