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1989 USAMO Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==
For each positive integer <math>n</math>, let
<div style="text-align:center;">
<math>S_n = 1 + \frac 12 + \frac 13 + \cdots + \frac 1n</math>


<math>T_n = S_1 + S_2 + S_3 + \cdots + S_n</math>
For each positive [[integer]] <math>n</math>, let
<cmath> \begin{align*}
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}.
\end{align*} </cmath>
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>.


<math>U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}</math>.
== Solution ==
</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>.


== Solution ==
We note that for all integers <math>n \ge 2</math>,
{{solution}}
<cmath>\begin{align*}
T_{n-1} &= 1 + \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*}</cmath>
 
It then follows that
<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 \\
&= \left(nS_n - n\right) + S_n - n = (n + 1)S_n - 2n .
\end{align*}</cmath>
 
If we let <math>n=1989</math>, we see that <math>(a,b,c,d) = (1989,1989,1990, 2\cdot 1989)</math> is a suitable solution.  <math>\blacksquare</math>
 
Notice that it is also possible to use induction to prove the equations relating <math>T_n</math> and <math>U_n</math> with <math>S_n</math>.
 
 
{{alternate solutions}}
 
== See Also ==


== See also ==
{{USAMO box|year=1989|before=First question|num-a=2}}
{{USAMO box|year=1989|before=First question|num-a=2}}
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356633#p356633 Discussion on AoPS/MathLinks]
{{MAA Notice}}


[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Algebra Problems]]

Latest revision as of 18:10, 18 July 2016

Problem

For each positive integer $n$, let \begin{align*} 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}. \end{align*} 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

We note that for all integers $n \ge 2$, \begin{align*} T_{n-1} &= 1 + \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*}

It then follows that \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*}

If we let $n=1989$, we see that $(a,b,c,d) = (1989,1989,1990, 2\cdot 1989)$ is a suitable solution. $\blacksquare$

Notice that it is also possible to use induction to prove the equations relating $T_n$ and $U_n$ with $S_n$.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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