Difference between revisions of "2005 PMWC Problems/Problem T6"

Revision as of 14:12, 9 October 2007

\begin{eqnarray*}
1+2 &=& 3 \\
4+5+6 &=& 7+8 \\
9+10+11+12 &=& 13+14+15 (Error compiling LaTeX. Unknown error_msg)

$\vdots$ If this pattern is continued, find the last number in the $80$ th row (e.g. the last number of the third row is $15$ ).

There are 3 numbers in the first row, 5 numbers in the second row, and $2n+1$ numbers in the $n$ th row. Thus for the $n$ th row, the last number is

$\left(\sum_{i=1}^{n} 2n+1\right)$ $= \left(\sum_{i=1}^n 2n-1\right) + 2n$ $= n^2 + 2n$

Since the sum of the first $n$ odd number is $n^2$ , the last number on the $80$ th row is $80^2 + 2\cdot 80 = 6560$ .

@@ Line 15: / Line 15: @@
 <cmath>= n^2 + 2n</cmath>
-Since the sum of the first <math>n</math> odd number is <math>n^2</math>. So the last number on the <math>80</math>th row is <math>80^2 + 2\cdot 80 = 6560</math>.
+Since the sum of the first <math>n</math> odd number is <math>n^2</math>, the last number on the <math>80</math>th row is <math>80^2 + 2\cdot 80 = 6560</math>.
 == See also ==