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

Revision as of 22:19, 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):

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

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

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.

@@ Line 10: / Line 10: @@
 '''Part a):'''
+<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>
+<math>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}</math>

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

Revision as of 22:19, 27 November 2023

Problem 9

Solution