What is the relationship between the some of the first n natural numbers and the number of unique pairs in n+1

Question

The sum of the first $n$ natural numbers is $\displaystyle\frac{n}{2}(n+1)$

$\displaystyle {n+1 \choose 2}$ is $\displaystyle\frac{n}{2}(n+1)$

Obviously they are equal, but how should I think about why they are equal. It seems there must be some logical explanation that I can't figure out.

Answer 1

Consider the numbers $1,2,3\dots n,n+1$ on the number line. How many segments have their endpoints on two of these points? Easy, there are $\binom{n+1}{2}$ ways to select the endpoints.

Another way to count it is the following. How many segments have length $1$? Well, if know the endpoint that is at the left we can uniquely determine the segment. Since there are $n$ options for that point (since it can be any number from $1$ to $n$), the answer is $n$.

How many segments have length $2$? The answer is $n-1$ since there are $n-1$ options for the endpoint at the left.

Following this line of thought the number of segments is $n+(n-1)+(n-2)\dots+2+1$

However when we counted it using the other method we got there where $\binom{n}{2}$. Since we counted the same thing we obtain $\binom{n+1}{2}=1+2+3+\dots +n$

Answer 2

Consider $\binom {n+1}2$ ways of choosing 2 balls from $n+1$ balls. Order the balls in some way.

If the first ball is chosen, then there are $n$ ways to choose next. If the first ball is not chosen and the second ball is chosen, then there are $n-1$ ways to choose next. $$\vdots$$ If all the first, second, ..., $(n-1)$th balls are not chosen and the $n$th ball is chosen, then there is 1 way to choose next.

$$\binom{n+1}2 = n+(n-1) + \cdots + 1$$

Answer 3

Think about a way to count the number of ways you can choose $2$ from $n+1$.

Take the $n+1$st object. You can pair it with any of the $n$ other objects. So far that makes $n$ ways to choose two objects.

You could also have a pair that includes the $n$th object. We have already counted the pair of the $n$ and $n+1$st object, but we can pair the $n$th object with any of the $n-1$ lower-numbered objects. We add $n-1$ new pairs this way.

And we can pair the $n-1$st object with any of the next $n-2$ objects to make $n-2$ new pairs.

As we look at each new object in turn and pair it with lower-numbered objects, the number of pairs we find adds up like this:

$$n + (n-1) + (n-2) + \cdots + + 2 + 1$$

where the final "$1$" is the pair formed from the $2$nd object and the $1$st, after which all objects have been paired once with every other object.

And that's the sum of the integers $1$ through $n$.