We have seen that the number of triangles in a complete graph with 7 vertices can be calculated using the pattern \((5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1=35\). This pattern gets very long for complete graphs with more vertices, but we have seen sums from 1 to a number before, and we had a shortcut! Recall from Example 12.8 that \(1+2+3+\cdots+(n-1)=\frac{n(n-1)}{2}\). This makes finding the sum from 1 up to \(n-1\) shorter. We can write each of the sums we need in the form \(\frac{n(n-1)}{2}\).

To find the sum from 1 to 5 , since \(n-1=5\), use \(n=6: 1+2+3+4+5=\frac{6(5)}{2}\)

To find the sum from 1 to 4 , since \(n-1=4\), use \(n=5: 1+2+3+4=\frac{5(4)}{2}\)

To find the sum from 1 to 3 , since \(n-1=3\), use \(n=4: 1+2+3=\frac{4(3)}{2}\)

To find the sum from 1 to 2 , since \(n-1=2\), use \(n=3: 1+2=\frac{3(2)}{2}\)

To find the sum from 1 to 1 , (Sounds silly, but it helps to see the pattern!) since \(n-1=1\), use \(n=2: 1=\frac{2(1)}{2}\) Use these to rewrite the calculation for the number of triangles in a graph with 7 vertices.

\[ \begin{aligned} & (5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1 \\ & =\underbrace{1}_{\frac{2(1)}{2}}+\underbrace{(1+2)}_{\frac{3(2)}{2}}+\underbrace{(1+2+3)}_{\frac{4(3)}{2}}+\underbrace{(1+2+3+4)}_{\frac{5(4)}{2}}+\underbrace{(1+2+3+4+5)}_{\frac{6(5)}{2}} \\ & =\frac{2(1)}{2}+\frac{3(2)}{2}+\frac{4(3)}{2}+\frac{5(4)}{2}+\frac{6(5)}{2} \\ & =\underbrace{\frac{1(2)+2(3)+3(4)+4(5)+5(6)}{2}}=35 \end{aligned} \]

The number of triangles in complete graph with 7 vertices.

Similarly, for the number of triangles in a complete graph with 8 vertices, instead of

\((6+5+4+3+2+1)+(5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+1=56\)

use the shorter formula \(\frac{1(2)+2(3)+3(4)+4(5)+5(6)+6(7)}{2}=56\).
Use this pattern to find the number of triangles in a complete graph with 10 vertices. In which row and entry of Pascal's triangle can you also find this result?