Simplify the expression $n + 2\binom{n}{2}$.

Question

Simplify the expression $n + 2\binom{n}{2}$. Try to prove your simplification using a counting argument.

I tried to substitute some numbers for n but I couldn't find a pattern. It would be great if someone could help me.

Answer 1

$$ n+2\binom{n}{2} = n + \frac{2n!}{2!(n-2)!} =n + \frac{n!}{(n-2)!} = n+(n)(n-1)=n^2 $$

Answer 2

Well to answer your question, $\binom{n}{2}$ is the number of boxes you can pick in the upper triangle of a $n^2$ square without its diagonal.

As an example, $\binom{4}{2}$ is like listing every couples of distincts elements from {1,2,3,4} without any order, so you can make {1,2}, {1,3}, {1,4}, {2,3}, {2,4} and {3,4}. It makes 6 couples, therefore $\binom{4}{2}$ is 6.

So $n$ is the number of boxes in the diagonal, $2\binom{n}{2}$ is all the boxes in the square without those on the diagonal. Therefore $n + 2\binom{n}{2} = n^2$ (the number of boxes in your square).

Answer 3

This is an old question, but I believe I have a clever solution...

Draw an $n$-gon. $\binom{n}{2}$ represents the number of sides and diagonals combined.

Now, determine a path to be a route along a line segment from any point A to point B. There are $2 \cdot \binom{n}{2}$ paths, since we have $\binom{n}{2}$ line segments, but you have two directions - one from point A to point B, and one from point B to point A.

However, if we choose any point on the $n$-gon, you can draw a loop back to itself, and there are $n$ of these, since there are $n$ points.

This is $\binom{n}{2}\cdot 2 + n$ in total.

Now think about it this way: at each point of the $n$-gon, it can go to any of the $n$ points (including itself!) on the perimeter of the $n$-gon. This goes for each of the $n$ points so it equals $n^2$.