2013 Mock AIME I Problems/Problem 10
Problem
Let 
denote the 
th triangular number, i.e. 
. Let 
and be relatively prime positive integers so that ![\[\sum_{i=3}^\infty \sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\dfrac{m}{n}.\]](http://latex.artofproblemsolving.com/2/a/a/2aa8b65deee59ad25cfb6da6637ab4d1b8878fb0.png)
Find 
.
Solution
Note ![\[\sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\frac{\frac{3}{T_i}}{1-\frac{3}{T_i}}=\frac{3}{T_i-3}.\]](http://latex.artofproblemsolving.com/6/8/3/683f3a764059807c3854b9e3bfb9ea2ef673c00c.png)
So, after using the formula for triangular numbers, we wish to evaluate ![\[\sum_{i=3}^\infty\frac{3}{T_i-3}=\sum_{i=3}^\infty\frac{6}{i^2+i-6}=\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}.\]](http://latex.artofproblemsolving.com/b/d/e/bde3959b81b0a3348e7cc71c6a2ec06c2f41573e.png)
By partial fraction decomposition, 
. Telescoping, we obtain ![\[\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\sum_{i=3}^\infty \frac{1}{i-2}-\frac{1}{i+3}\right)=\frac{6}{5}(1+\tfrac12+\tfrac13+\tfrac14+\tfrac15)=\frac{137}{50}.\]](http://latex.artofproblemsolving.com/d/b/6/db6219ca83315dc521314f10b28a1ca60637bcde.png)
Hence, 
.
