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

## Problem

$\begin{eqnarray*} 1+2 &=& 3 \\ 4+5+6 &=& 7+8 \\ 9+10+11+12 &=& 13+14+15\end{eqnarray*}$ $$\vdots$$ 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 (We use the fact that the sum of the first $n$ odd numbers is $n^2$):

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

The last number on the $80$th row is $80^2 + 2\cdot 80 = 6560$.

## See also

 2005 PMWC (Problems) Preceded byProblem T5 Followed byProblem 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
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