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

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</math>
 
</math>
  
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== Solution ==
+
== Solution 1 ==
 
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>.
−
Now <math>\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) }}</math>.
+
 
 +
Now <math>\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}}.</math>
 +
 
 +
Note: For the sake of completeness, I put the full derivation of the partial fraction decomposition of <math>\frac{2}{n(n+1)}</math> 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 <math>\sum_{i=1}^{n} \frac{1}{t_n}.</math>
 +
 
 +
With some guess and check:
 +
 
 +
<math>n=1</math>
 +
 
 +
\begin{align*}
 +
\frac{1}{t_1}&=\frac{2}{1(2)} \\
 +
&=\frac{2}{2} \\
 +
\end{align*}
 +
 
 +
<math>n=2</math>
 +
 
 +
\begin{align*}
 +
\frac{1}{t_1}+\frac{1}{t_2}&=\frac{2}{2}+\frac{2}{2(3)} \\
 +
&=\frac{4}{3} \\
 +
\end{align*}
 +
 
 +
<math>n=3</math>
 +
 
 +
\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 <math>\frac{2}{2}, \frac{4}{3},</math> and <math>\frac{6}{4},</math> we can conjecture <math>\sum_{i=1}^{n} \frac{1}{t_n} = \frac{2n}{n+1}.</math> A quick check shows that for <math>n=4</math> gives <math>\frac{8}{5},</math> which means our inductive hypothesis is most likely correct. Thus, our answer is <math>\sum_{i=1}^{2002} \frac{1}{t_n} = \frac{2(2002)}{2002+1}=\boxed{\textbf{(C) } \frac{4004}{2003}}.</math>
 +
 
 +
We will prove this with induction for all <math>n \geq 1</math> as promised.
 +
 
 +
Base case: <math>n=1.</math>
 +
 
 +
\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 <math>1=1,</math> the base case is proven.
 +
 
 +
Induction step: <math>n=k+1.</math>
 +
 
 +
Assume: <math>n=k</math> <math>\implies</math> <math>\sum_{i=1}^{k} \frac{1}{t_k} = \frac{2k}{k+1}.</math>
 +
 
 +
We want to prove: <math>n=k+1</math> <math>\implies</math> <math>\sum_{i=1}^{k+1} \frac{1}{t_k} = \frac{2(k+1)}{k+2}.</math>
 +
 
 +
We are given <math>\sum_{i=1}^{k} \frac{1}{t_k} = \frac{2k}{k+1}.</math> Add <math>\frac{1}{t_{k+1}}</math> 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 ==
 
== 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}}

Latest revision as of 09:05, 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
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All AMC 12 Problems and Solutions

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