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Difference between revisions of "1970 Canadian MO Problems/Problem 9"

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== Problem ==
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== Problem 9 ==
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Let <math>f(n)</math> be the sum of the first <math>n</math> terms of the sequence
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<cmath> 0, 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, \ldots\, .  </cmath>
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a) Give a formula for <math>f(n)</math>.
  
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Let <math>f(n)</math> be the sum of the first <math>n</math> terms of the sequence 0, 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, \ldots\, . a) Give a formula for <math>f(n)</math>.
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b) Prove that <math>f(s+t)-f(s-t)=st</math> where <math>s</math> and <math>t</math> are positive integers and <math>s>t</math>.
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b) Prove that <math>f(s+t)-f(s-t)=s</math>t where <math>s</math> and <math>t</math> are positive integers and <math>s>t</math>.
 
  
 
== Solution ==
 
== Solution ==
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'''Part a):'''
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<math>f(n)=\begin{cases} 2\sum_{k=1}^{\frac{n-1}{2}}k,\; & n\;is\;odd \\ 2\sum_{k=1}^{\frac{n}{2}}k-\frac{n}{2},\; & n\;is\;even\end{cases}</math>
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<math>f(n)=\begin{cases} \left( \frac{n-1}{2} \right)\left( \frac{n-1}{2}+1 \right),\; & n\;is\;odd \\
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        \left( \frac{n}{2} \right)\left( \frac{n}{2}+1 \right)-\frac{n}{2},\; & n\;is\;even\end{cases}</math>
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<math>f(n)=\begin{cases} \frac{n^2-1}{4},\; & n\;is\;odd \\
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                    \frac{n^2}{4},\; & n\;is\;even\end{cases}</math>
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Using <math>(-1)^n</math> in the formula we can have:
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<math>\frac{(-1)^n-1}{2}=\begin{cases} -1,\; & n\;is\;odd \\
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                    0,\; & n\;is\;even\end{cases}</math>
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Therefore,
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<math>f(n)=\frac{n^2+\frac{(-1)^n-1}{2}}{4}</math>
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<math>f(n)=\frac{2n^2-1+(-1)^n}{8}</math>
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'''Part b):'''
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<math>f(s+t)-f(s-t)=\frac{2(s+t)^2-1+(-1)^{s+t}}{8}-\frac{2(s-t)^2-1+(-1)^{s-t}}{8}</math>
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since <math>(s+t)\equiv (s-t)\;(mod\;2)</math>, then <math>(-1)^{s+t}=(-1)^{s-t}</math> and our expression reduces to:
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<math>f(s+t)-f(s-t)=\frac{2(s+t)^2-2(s-t)^2}{8}=\frac{(s+t)^2-(s-t)^2}{4}</math>
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<math>f(s+t)-f(s-t)=\frac{s^2+2st+t^2-s^2+2st-t^2}{4}=\frac{4st}{4}=st</math>
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Tomas Diaz. orders@tomasdiaz.com
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{{alternate solutions}}

Latest revision as of 22:31, 27 November 2023

Problem 9

Let $f(n)$ be the sum of the first $n$ terms of the sequence \[0, 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, \ldots\, .\] a) Give a formula for $f(n)$.

b) Prove that $f(s+t)-f(s-t)=st$ where $s$ and $t$ are positive integers and $s>t$.

Solution

Part a):

$f(n)=\begin{cases} 2\sum_{k=1}^{\frac{n-1}{2}}k,\; & n\;is\;odd \\ 2\sum_{k=1}^{\frac{n}{2}}k-\frac{n}{2},\; & n\;is\;even\end{cases}$

$f(n)=\begin{cases} \left( \frac{n-1}{2} \right)\left( \frac{n-1}{2}+1 \right),\; & n\;is\;odd \\ \left( \frac{n}{2} \right)\left( \frac{n}{2}+1 \right)-\frac{n}{2},\; & n\;is\;even\end{cases}$

$f(n)=\begin{cases} \frac{n^2-1}{4},\; & n\;is\;odd \\ \frac{n^2}{4},\; & n\;is\;even\end{cases}$

Using $(-1)^n$ in the formula we can have:

$\frac{(-1)^n-1}{2}=\begin{cases} -1,\; & n\;is\;odd \\ 0,\; & n\;is\;even\end{cases}$

Therefore,

$f(n)=\frac{n^2+\frac{(-1)^n-1}{2}}{4}$

$f(n)=\frac{2n^2-1+(-1)^n}{8}$

Part b):

$f(s+t)-f(s-t)=\frac{2(s+t)^2-1+(-1)^{s+t}}{8}-\frac{2(s-t)^2-1+(-1)^{s-t}}{8}$

since $(s+t)\equiv (s-t)\;(mod\;2)$, then $(-1)^{s+t}=(-1)^{s-t}$ and our expression reduces to:

$f(s+t)-f(s-t)=\frac{2(s+t)^2-2(s-t)^2}{8}=\frac{(s+t)^2-(s-t)^2}{4}$

$f(s+t)-f(s-t)=\frac{s^2+2st+t^2-s^2+2st-t^2}{4}=\frac{4st}{4}=st$

Tomas Diaz. orders@tomasdiaz.com

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

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