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

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 (ProblemsAnswer KeyResources)
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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