Triangular pyramid. Find the sum of certain numbers on pyramid.

Question

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.

Answer 1

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 $(x-1)^2+k$

Answer 2

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 $1\leq k \leq 2n-1$.