Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 Mock AIME I Problems/Problem 10"

m (formatted sections)
m (links, formatting changes)
 
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
−
Note <cmath>\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}.</cmath> So we wish to evaluate <cmath>\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)}.</cmath> It is not difficult to check <math>\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\frac{1}{i-2}-\frac{1}{i+3}\right)</math>. Telescoping, we obtain <cmath>\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+1/2+1/3+1/4+1/5)=\frac{137}{50}.</cmath> Hence, <math>m+n=\boxed{187}</math>.
+
Note <cmath>\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}.</cmath> So, after using the formula for [[triangular numbers#Formula|triangular numbers]], we wish to evaluate <cmath>\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)}.</cmath> By [[partial fraction decomposition]], <math>\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\frac{1}{i-2}-\frac{1}{i+3}\right)</math>. [[telescoping series|Telescoping]], we obtain <cmath>\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}.</cmath> Hence, <math>m+n=\boxed{187}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 13:38, 30 July 2024

Problem

Let $T_n$ denote the $n$th triangular number, i.e. $T_n=1+2+3+\cdots+n$. Let $m$ and $n$ be relatively prime positive integers so that \[\sum_{i=3}^\infty \sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\dfrac{m}{n}.\] Find $m+n$.


Solution

Note \[\sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\frac{\frac{3}{T_i}}{1-\frac{3}{T_i}}=\frac{3}{T_i-3}.\] So, after using the formula for triangular numbers, we wish to evaluate \[\sum_{i=3}^\infty\frac{3}{T_i-3}=\sum_{i=3}^\infty\frac{6}{i^2+i-6}=\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}.\] By partial fraction decomposition, $\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\frac{1}{i-2}-\frac{1}{i+3}\right)$. Telescoping, we obtain \[\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\sum_{i=3}^\infty \frac{1}{i-2}-\frac{1}{i+3}\right)=\frac{6}{5}(1+\tfrac12+\tfrac13+\tfrac14+\tfrac15)=\frac{137}{50}.\] Hence, $m+n=\boxed{187}$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_Mock_AIME_I_Problems/Problem_10&oldid=223995"