2005 PMWC Problems/Problem T6

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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 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
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