Difference between revisions of "2013 Mock AIME I Problems/Problem 10"
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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>. | 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>. | ||
| + | |||
| + | == See also == | ||
| + | * [[2013 Mock AIME I Problems]] | ||
| + | * [[2013 Mock AIME I Problems/Problem 9|Preceded by Problem 9]] | ||
| + | * [[2013 Mock AIME I Problems/Problem 11|Followed by Problem 11]] | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 13:30, 30 July 2024
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 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)
It is not difficult to check 
. Telescoping, we obtain ![\[\sum_{i=3}^\infty\frac{6}{(i-2)(i+3)}=\frac{6}{5}\left(\sum_{i=3}^\infty \frac{1}{i-2}-\frac{1}{i+3}\right)=\frac{6}{5}(1+1/2+1/3+1/4+1/5)=\frac{137}{50}.\]](http://latex.artofproblemsolving.com/4/f/4/4f461d52b7532f750f4e9353ae5d4579df9b11c7.png)
Hence, 
.
