2002 AMC 12P Problems/Problem 11

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 solution 1, $\frac{1}{t_n} = \frac{2}{n(n+1)}.$ However, knowing that $2003$ is prime,

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 12 Problems and Solutions

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2002 AMC 12P Problems/Problem 11

Contents

Problem

Solution 1

Solution 2 (Cheese)

See also