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Difference between revisions of "2002 AMC 12P Problems/Problem 11"

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== Solution 2 (Cheese) ==
 
== Solution 2 (Cheese) ==
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As with solution 1, <math>\frac{1}{t_n} = \frac{2}{n(n+1)}.</math> However, knowing that <math>2003</math> is prime,
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As with all telescoping problems, there is a solution that involves induction. In competition, it is sufficient to conjecture the formula but not prove it. For sake of completeness, we will prove the formula.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=10|num-a=12}}
 
{{AMC12 box|year=2002|ab=P|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:19, 15 July 2024

Problem

Let $t_n = \frac{n(n+1)}{2}$ be the $n$th triangular number. Find

\[\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + ... + \frac{1}{t_{2002}}\]

$\text{(A) }\frac {4003}{2003} \qquad \text{(B) }\frac {2001}{1001} \qquad \text{(C) }\frac {4004}{2003} \qquad \text{(D) }\frac {4001}{2001} \qquad \text{(E) }2$

Solution 1

We may write $\frac{1}{t_n}$ as $\frac{2}{n(n+1)}$ and do a partial fraction decomposition. Assume $\frac{2}{n(n+1)} = \frac{A_1}{n} + \frac{A_2}{n+1}$.

Multiplying both sides by $n(n+1)$ gives $2 = A_1(n+1) + A_2(n) = (A_1 + A_2)n + A_1$.

Equating coefficients gives $A_1 = 2$ and $A_1 + A_2 = 0$, so $A_2 = -2$. Therefore, $\frac{2}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$.

Now $\frac{1}{t_1} + \frac{1}{t_2} + ... + \frac{1}{t_{2002}} = 2((\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{2002} - \frac{1}{2003})) = 2(1 - \frac{1}{2003}) = \boxed{\frac {4004}{2003} \text{(C) }}$.

Note: For the sake of completeness, I put the full derivation of the partial fraction decomposition of $\frac{2}{n(n+1)}$ here. However, on the contest, the decomposition step would be much faster since it is so well-known.

Solution 2 (Cheese)

As with all telescoping problems, there is a solution that involves induction. In competition, it is sufficient to conjecture the formula but not prove it. For sake of completeness, we will prove the formula.

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

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