2007 UNCO Math Contest II Problems/Problem 3: Difference between revisions

Latest revision as of 04:05, 12 January 2019

Problem

State the general rule illustrated here and prove it:

$1 ,\quad \begin{tabular}{cc} 1&1\\1&2\end{tabular} ,\quad \begin{tabular}{ccc} 1&1&1\\1&2&2\\1&2&3\end{tabular},\quad \begin{tabular}{cccc} 1&1&1&1\\1&2&2&2\\1&2&3&3\\1&2&3&4 \end{tabular} ,\quad \begin{tabular}{ccccc} 1&1&1&1&1\\1&2&2&2&2\\1&2&3&3&3\\1&2&3&4&4\\1&2&3&4&5 \end{tabular} ,\quad \cdots$

Solution

$1^2+2^2+3^2+\cdots+n^2=1\cdot{n}+3\cdot{(n-1)}+5\cdot{(n-2)}+\cdots+(2n-1)\cdot{1}$

@@ Line 8: / Line 8: @@
 == Solution ==
-{{solution}}
+<math>1^2+2^2+3^2+\cdots+n^2=1\cdot{n}+3\cdot{(n-1)}+5\cdot{(n-2)}+\cdots+(2n-1)\cdot{1}</math>
 == See Also ==

2007 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
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2007 UNCO Math Contest II Problems/Problem 3: Difference between revisions

Latest revision as of 04:05, 12 January 2019

Problem

Solution

See Also