Difference between revisions of "2005 PMWC Problems/Problem T6"
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== Solution == | == Solution == | ||
− | There are 3 numbers in the first row, 5 numbers in the second row, and <math>2n+1</math> numbers in the <math>n</math>th row. Thus for the <math>n</math>th row, the last number is | + | There are 3 numbers in the first row, 5 numbers in the second row, and <math>2n+1</math> numbers in the <math>n</math>th row. Thus for the <math>n</math>th row, the last number is (We use the fact that the sum of the first <math>n</math> odd numbers is <math>n^2</math>): |
<cmath>\left(\sum_{i=1}^{n} 2n+1\right)</cmath> | <cmath>\left(\sum_{i=1}^{n} 2n+1\right)</cmath> | ||
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<cmath>= n^2 + 2n</cmath> | <cmath>= n^2 + 2n</cmath> | ||
− | + | 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 15:16, 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)
If this pattern is continued, find the last number in the th row (e.g. the last number of the third row is ).
Solution
There are 3 numbers in the first row, 5 numbers in the second row, and numbers in the th row. Thus for the th row, the last number is (We use the fact that the sum of the first odd numbers is ):
The last number on the th row is .
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 |