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Given that the set of natural numbers continue in the triangular pattern shown below, find the sum of the 6th number in row 7 and the 9th number in row 10.

Note: Row 1 is the first row of the triangle.

$$1$$ $$2\space\space3\space\space4$$ $$5\space\space6\space\space7\space\space8\space\space9$$ $$10\space\space11\space\space12\space\space13\space\space14\space\space15\space\space16$$ $$...$$

My approach (horrible): I found the quadratic sequence $x^2-x+1$ which produced the median of each row. Keeping the median as a reference point, I then counted manually to find the numbers in each row.

Ian L
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2 Answers2

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Your approach is quite reasonable. You can improve it by replacing the manual count with the recognition that there are $2x-1$ numbers in row $x$, so $x-1$ to the left of the median and $x-1$ to the right. You can use this to figure out the change from the median for a given location.

A simpler approach is to note that row $x$ has $x^2$ as its last number, so $(x-1)^2+1$ as its first. The $k^{th}$ number in row $x$ is then $(x-1)^2+k$

Ross Millikan
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The last number in $n$-th row is $n^2$.

It's justified since $1+3+\ldots+(2n-1)=n^2$

Hence, the middle number in $n$-th row is $(n-1)^2+n=n^2-n+1$.

The $k$-th number in $n$-th row is $(n-1)^2+k$ where $1\leq k \leq 2n-1$.

Ng Chung Tak
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