Difference between revisions of "1989 USAMO Problems/Problem 1"

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== Problem ==
 
== Problem ==
For each positive integer <math>n</math>, let
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For each positive [[integer]] <math>n</math>, let
 
<div style="text-align:center;">
 
<div style="text-align:center;">
 
<math>S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n</math>
 
<math>S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n</math>
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<math>U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}</math>.
 
<math>U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}</math>.
 
</div>
 
</div>
Find, with proof, integers <math>0 < a,\ b,\ c,\ d < 1000000</math> such that <math>\displaystyle T_{1988} = a S_{1989} - b</math> and <math>\displaystyle U_{1988} = c S_{1989} - d</math>.
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Find, with proof, integers <math>0 < a,\ b,\ c,\ d < 1000000</math> such that <math>T_{1988} = a S_{1989} - b</math> and <math>U_{1988} = c S_{1989} - d</math>.
  
 
== Solution ==
 
== Solution ==
Let us prove that <math>\displaystyle T_{n-1} = nS_n - n</math>. Expanding:
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If we re-group the terms of <math>T_{n-1}</math>,
  
:<math>\left(1\right) + \left(1 + \frac 12\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1n\right) = n\left(\sum_{i=1}^n \frac 1i\right) - n \displaystyle</math>
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<cmath>\begin{align*}
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T_{n-1} &= \left(1\right) + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\
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&= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\
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&= n \cdot S_{n} - n\end{align*}</cmath>
  
Grouping like terms, there are <math>n-1</math> <math>\displaystyle 1</math>s, <math>n-2</math> <math>\frac 12</math>s, and so on:
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Thus, for <math>n = 1989</math>, <math>T_{1988} = 1989 S_{1989} - 1989 \Longrightarrow \boxed{a = b = 1989}</math>.
  
:<math>\left(\sum_{i=1}^{n-1} \frac 1i \cdot (n - i)\right) = n\left(\sum_{i=1}^n \frac 1i\right) - n</math>
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For the second part, applying this result gives
:<math>\left(\sum_{I=1}^{n-1} \frac ni\right) - (n - 1)  =\left(\sum_{i=1}^n \frac ni\right) - n</math>
 
:<math>-(n-1) = n \cdot \frac 1n - n \Longrightarrow 1 - n = 1 - n</math>
 
  
which completes our proof. Thus, for <math>\displaystyle n = 1989</math>, we have that <math>\displaystyle T_{1988} = 1989 S_{1989} - 1989</math>, and so <math>a = b = 1989 \displaystyle</math>.
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<cmath>\begin{align*}U_{n-1} &= \sum_{i=2}^{n} \frac{T_{i-1}}{i} = \sum_{i=2}^{n}\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1\\
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&= \left(nS_n - n\right) + S_n - n = (n + 1)S_n - 2n\end{align*}</cmath>
  
For the second part, use our previously derived identity to determine <math>\displaystyle U_{n-1}</math> in terms of <math>\displaystyle S_n</math>. The problem simplifies to:
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For <math>n = 1989</math>, we get that <math>\boxed{c = 1990, d = 2 \cdot 1989 = 3978}</math>.
 
 
:<math>U_{n-1} = \frac{2 S_2 - 2}{2} +  \frac{3 S_3 - 3}{3}  + \ldots + \frac{nS_{n} - n}{n}</math>
 
:<math>=S_2 + S_3 + \ldots S_{n} - (n - 1) + \left(S_1 - S_1\right)</math>
 
:<math>\displaystyle = T_{n-1} + S_n - (n-1)</math>
 
:<math>\displaystyle = \left(nS_n - n\right) + S_n - n + 1 - S_1</math>
 
:<math>\displaystyle = (n + 1)S_n - 2n</math>
 
 
 
Thus, we have <math>\displaystyle U_{n-1} = (n+1)S_n - 2n</math>. For <math>\displaystyle n = 1989</math>, we get that <math>\displaystyle c = 1990</math> and <math>d = 2 \cdot 1989 = 3978</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 20:58, 10 January 2008

Problem

For each positive integer $n$, let

$S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n$

$T_n = S_1 + S_2 + S_3 + \cdots + S_n$

$U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}$.

Find, with proof, integers $0 < a,\ b,\ c,\ d < 1000000$ such that $T_{1988} = a S_{1989} - b$ and $U_{1988} = c S_{1989} - d$.

Solution

If we re-group the terms of $T_{n-1}$,

\begin{align*} T_{n-1} &= \left(1\right) + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\  &= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\ &= n \cdot S_{n} - n\end{align*}

Thus, for $n = 1989$, $T_{1988} = 1989 S_{1989} - 1989 \Longrightarrow \boxed{a = b = 1989}$.

For the second part, applying this result gives

\begin{align*}U_{n-1} &= \sum_{i=2}^{n} \frac{T_{i-1}}{i} = \sum_{i=2}^{n}\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1\\ &= \left(nS_n - n\right) + S_n - n = (n + 1)S_n - 2n\end{align*}

For $n = 1989$, we get that $\boxed{c = 1990, d = 2 \cdot 1989 = 3978}$.

See also

1989 USAMO (ProblemsResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions
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