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

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Problem: Let <math>T_n</math> denote the <math>n</math>th triangular number, i.e. <math>T_n=1+2+3+\cdots+n</math>. Let <math>m</math> and <math>n</math> be relatively prime positive integers so that <cmath>\sum_{i=3}^\infty \sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\dfrac{m}{n}.</cmath> Find <math>m+n</math>.  
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== Problem ==
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Let <math>T_n</math> denote the <math>n</math>th triangular number, i.e. <math>T_n=1+2+3+\cdots+n</math>. Let <math>m</math> and <math>n</math> be relatively prime positive integers so that <cmath>\sum_{i=3}^\infty \sum_{k=1}^\infty \left(\dfrac{3}{T_i}\right)^k=\dfrac{m}{n}.</cmath> Find <math>m+n</math>.  
  
  
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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>.
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== Solution ==
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 13:32, 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 we wish to evaluate \[\sum_{i=3}^\infty\frac{3}{T_i-3}=\sum_{i=3}^\infty\frac{6}{i^2+i-6}=\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}.\] It is not difficult to check $\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\frac{1}{i-2}-\frac{1}{i+3}\right)$. Telescoping, we obtain \[\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\sum_{i=3}^\infty \frac{1}{i-2}-\frac{1}{i+3}\right)=\frac{6}{5}(1+1/2+1/3+1/4+1/5)=\frac{137}{50}.\] Hence, $m+n=\boxed{187}$.

See also

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