How to prove $\binom{n+2}{2}-\binom{n-1}{2}=3n$? While reading a solution to one of the problems in the textbook, I found this in one of the steps of solution:

$\binom{n+2}{2}-\binom{n-1}{2}=3n$

I was puzzled about how that result was brought. I thought I can too come up with that with simplification. So I tried simplifying, but out of luck. Then I tried for some values of $n$ and realized that it is indeed correct. Is this something stupid easy and I miss something basic? How do I prove this? I will prefer both solutions: by algebraic simplification and by double counting argument (if there can be any).
 A: $\binom n2$ is the $(n-1)$-th triangular number, that is
$$\binom n2=1+2+\cdots+(n-1).$$
Then
$$\binom{n+2}2=1+2+\cdots+(n+1)$$
and
$$\binom{n-1}2=1+2+\cdots+(n-2).$$
What's left over when we subtract these?
A: A combinatorial proof. 
$\binom{n+2}{2}$ is the number unordered pairs of distinct elements of $\{1,2,\dots,n+2\}$. Of those, 


*

*$\binom{n-1}{2}$ are pairs of elements both in $\{1,2,\dots,n-1\}$,

*$3(n-1)$ are pairs where one element is in $\{n,n+1,n+2\}$ and one is an element of $\{1,\dots,n-1\}$. 

*$3=\binom{3}{1}$ are pairs where both are in $\{n,n+1,n+2\}$. 


So $$\binom{n+2}{2}=\binom{n-1}{2}+3(n-1)+3$$
which gives your result.
More generally, by the same argument, we have that:
$$\binom{a+b}{2} = \binom{a}{2}+ab+\binom{b}{2}$$
In particluar, if $b=2k+1$ and $a=n-k$ then:
$$\binom{n+k+1}{2}=\binom{n-k}{2}+(2k+1)n$$
A: In general you have ${m \choose 2} = m(m-1)/2$.  So you can rewrite this as
$$ {(n+2)(n+1) \over 2} - {(n-1)(n-2) \over 2} $$
and if you expand out the numerator and denominator you get
$$ {(n^2+3n+2) - (n^2-3n+2) \over 2} = {6n \over 2} = {3n}. $$
As for a combinatorial argument, the left-hand side counts subsets of $\{1, 2, \ldots, n+2 \}$, of size 2, which are not subsets of $\{1 ,2, \ldots n-1\}$.   That is, these are subsets of $\{ 1, 2, \ldots, n+2 \}$ which contain at least one of $n, n+1$ and $n+2$.  There are three such sets containing two of $n, n+1$, and $n+2$, namely $\{n, n+1\}, \{n, n+2\}$ and $\{n+1, n+2\}$.  And there are $3(n-1)$ which contain one of $n, n+1$, and $n+2$ - to construct such a set you need to pick one of those three "special" elements and one of $1, 2, \ldots, n-1$.  Adding those together you get $3n$.
A: Hint:
\begin{align}
\binom{n+2}{2}-\binom{n-1}{2} = \frac{(n+2)(n+1)}{2}-\frac{(n-1)(n-2)}{2} 
\end{align}
A: $$\binom{n+2}{2} - \binom{n-1}{2} = \dfrac{(n+2)(n+1)}{2} - \dfrac{(n-1)(n-2)}{2} = \dfrac{n^2 + 3n + 2}{2} - \dfrac{n^2 - 3n + 2}{2} = 3n.$$
A: $\binom{n+2}{2}-\binom{n-1}{2}=3n$ Here's how to do this:
$\frac{(n+2)!}{2n!}-\frac{(n-1)!}{2(n-3)!}$
then
$\frac{(n+2)(n+1)}{2}-\frac{(n-1)(n-2)}{2}$
after solving this you get $3n$ as the answer.
A: You can prove it using generating functions.
Note that
$$
\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n.
$$
Thus
$$
\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}=\sum_{n=0}^\infty \left[\binom{n+2}{2}-\binom{n-1}{2}\right]x^n
$$
while
$$
\frac{1}{(1-x)^3}-\frac{x^3}{(1-x)^3}
=\frac{1+x+x^2}{(1-x)^2}
=\sum_{n=0}^\infty [(n+1)+n+(n-1)]x^n
=\sum_{n=0}^\infty (3n)x^n.
$$
Hence the identity follows.
A: Recall that ${k \choose 2} = \frac{k(k-1)}{2}$.  Using this formula, we get that
${n+2 \choose 2} - {n-1 \choose 2} = \frac{(n+2)(n+1)}{2} - \frac{(n-1)(n-2)}{2} = \frac{n^2 + 3n + 2}{2 } - \frac{n^2 - 3n + 2}{2}$
$ = \frac{6n}{2} = 3n.$
