Difference between revisions of "2002 AMC 12P Problems/Problem 11"
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== Solution == | == Solution == | ||
We may write <math>\frac{1}{t_n}</math> as <math>\frac{2}{n(n+1)}</math> and do a partial fraction decomposition. | We may write <math>\frac{1}{t_n}</math> as <math>\frac{2}{n(n+1)}</math> and do a partial fraction decomposition. | ||
| − | Assume <math>\frac{2}{n(n+1)} = \frac{A_1}{n} + \frac{A_2}{n+1}</math>. Multiplying both sides by | + | Assume <math>\frac{2}{n(n+1)} = \frac{A_1}{n} + \frac{A_2}{n+1}</math>. |
| − | <math>n(n+1)</math> gives <math>2 = A_1(n+1) + A_2(n) = (A_1 + A_2)n + A_1</math>. | + | Multiplying both sides by <math>n(n+1)</math> gives <math>2 = A_1(n+1) + A_2(n) = (A_1 + A_2)n + A_1</math>. |
Equating coefficients gives <math>A_1 = 2</math> and <math>A_1 + A_2 = 0</math>, so <math>A_2 = -2</math>. Therefore, <math>\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}</math>. | Equating coefficients gives <math>A_1 = 2</math> and <math>A_1 + A_2 = 0</math>, so <math>A_2 = -2</math>. Therefore, <math>\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}</math>. | ||
Revision as of 13:51, 10 March 2024
Problem
Let 
be the 
th triangular number. Find
![\[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + ... + \frac{1}{t_{2002}}\]](http://latex.artofproblemsolving.com/9/a/3/9a334b618212e6acf0eb3eee2eb52e3997b96d39.png)
Solution
We may write 
as 
and do a partial fraction decomposition.
Assume 
.
Multiplying both sides by 
gives 
.
Equating coefficients gives 
and 
, so 
. Therefore, 
.
Now 
.
See also
| 2002 AMC 12P (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
