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

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Latest revision as of 10:05, 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 $\frac{1}{t_1} + \frac{1}{t_2} + ... + \frac{1}{t_{2002}} = 2((\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ... + (\frac{1}{2002} - \frac{1}{2003})) = 2(1 - \frac{1}{2003}) = \boxed{\textbf{(C) } \frac{4004}{2003}}.$

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 and practice, we will prove the formula for $\sum_{i=1}^{n} \frac{1}{t_n}.$

With some guess and check:

$n=1$

\begin{align*} \frac{1}{t_1}&=\frac{2}{1(2)} \\ &=\frac{2}{2} \\ \end{align*}

$n=2$

\begin{align*} \frac{1}{t_1}+\frac{1}{t_2}&=\frac{2}{2}+\frac{2}{2(3)} \\ &=\frac{4}{3} \\ \end{align*}

$n=3$

\begin{align*} \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}&=\frac{4}{3}+\frac{2}{3(4)} \\ =\frac{6}{4} \\ \end{align*}

From $\frac{2}{2}, \frac{4}{3},$ and $\frac{6}{4},$ we can conjecture $\sum_{i=1}^{n} \frac{1}{t_n} = \frac{2n}{n+1}.$ A quick check shows that for $n=4$ gives $\frac{8}{5},$ which means our inductive hypothesis is most likely correct. Thus, our answer is $\sum_{i=1}^{2002} \frac{1}{t_n} = \frac{2(2002)}{2002+1}=\boxed{\textbf{(C) } \frac{4004}{2003}}.$

We will prove this with induction for all $n \geq 1$ as promised.

Base case: $n=1.$

\begin{align*} \sum_{i=1}^{1} \frac{1}{t_n}&=\frac{1}{t_n} \\ &=\frac{2}{1(2)} \\ &=1 \\ &=\frac{2(1)}{1+1} \\ &=1 \\ \end{align*}

Since $1=1,$ the base case is proven.

Induction step: $n=k+1.$

Assume: $n=k$ $\implies$ $\sum_{i=1}^{k} \frac{1}{t_k} = \frac{2k}{k+1}.$

We want to prove: $n=k+1$ $\implies$ $\sum_{i=1}^{k+1} \frac{1}{t_k} = \frac{2(k+1)}{k+2}.$

We are given $\sum_{i=1}^{k} \frac{1}{t_k} = \frac{2k}{k+1}.$ Add $\frac{1}{t_{k+1}}$ on both sides and simplify.

\begin{align*} \sum_{i=1}^{k} \frac{1}{t_k} + \frac{1}{t_{k+1}} &= \frac{2k}{k+1} + \frac{1}{t_{k+1}} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2k}{k+1} + \frac{2}{(k+1)(k+2)} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2k(k+2)}{(k+1)(k+2)} + \frac{2}{(k+1)(k+2)} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2k(k+2)+2}{(k+1)(k+2)} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2k^2+4k+2}{(k+1)(k+2)} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2(k+1)^2}{(k+1)(k+2)} \\ \sum_{i=1}^{k+1} \frac{1}{t_k} &= \frac{2(k+1)}{(k+2)} \\ \end{align*}

Given the inductive hypothesis, we have proven the induction step. Thus, we have completed our proof for induction.

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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