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

m (Problem)
m (Solution)
Line 15: Line 15:
 
<cmath>= n^2 + 2n</cmath>
 
<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 ==
 
== See also ==

Revision as of 15:12, 9 October 2007

Problem

\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 $15$).

Solution

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 $80^2 + 2\cdot 80 = 6560$.

See also

2005 PMWC (Problems)
Preceded by
Problem T5
Followed by
Problem T7
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10