Simplify triangular sum of triangular numbers: $\sum_{i=1}^{n}(\frac12i(i+1))$ I'd like to simplify this expression, which sums up the first $n$ triangular numbers:
$$\sum_{i=1}^{n}(\frac12i(i+1))$$
which is equal to:
$$\sum_{i=0}^{n}((n-i)(i+1))$$
 A: In general we have:$$\sum_{i=r}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$
This is a useful equality that can be proved with induction on $n$.
The induction step is (triangle of Pascal): $$\binom{n+1}{r+1}+\binom{n+1}{r}=\binom{n+2}{r+1}$$
You can apply this to find: $$\sum_{i=1}^n\frac{i(i+1)}2=\sum_{i=1}^{n}\binom{i+1}{2}=\sum_{i=2}^{n+1}\binom{i}{2}=\binom{n+2}{3}$$
A: It is possible to give a direct combinatorial proof that:
$$\sum_{i=1}^{n-2}T_i=\binom n3$$
which follows the logic of this answer on a related question. https://math.stackexchange.com/a/2478631/627744
Suppose $n$ people wish to take every possible 3-person selfie. By definition, there are $\binom n3$ ways to do this.
Alternatively, imagine the $n$ people are all standing in a line. First compose all the selfies in which person 1 participates. Person 1 follows as person 2 walks down the line taking a pic with each of the remaining $n-2$ people. Then person 2 steps aside so that person 1 can follow person 3, taking all $n-3$ 3-person pics that include persons 1 and 3, but not 2. Then person 3 steps aside so that person 1 can take the $n-4$ pics that include person 4, but not persons 2 or 3. This process continues so that there are $T_{n-2}$ selfies with person 1.
Person 1 now steps aside permanently and we compose all the selfies in which person 2 participates, but person 1 doesn't. Person 2 now plays the previous role of person 1, following each person down the row, taking $T_{n-3}$ selfies in total.
In this way, $\sum_{i=1}^{n-2}T_i$ counts the total number of 3-person selfies.
A: Observe that
$$
\frac12i(i+1)+\frac16=\frac16(i+1)^3-\frac16i^3
$$ Then you have a telescoping sum giving
$$
\sum_{i=1}^n\frac12i(i+1)+\frac n6=\frac16(n+1)^3-\frac16.
$$
A: A good way is to simplify the summation as follows:
$$\sum_{i=1}^n \frac{i(i+1)}{2}=\frac{1}{2}\left[\sum_{i=1}^n i^2 + \sum_{i=1}^n i \right]$$
and then use the standard results for $\sum_\limits{i=1}^n i^2$ and $\sum_\limits{i=1}^n i$.
A: Obviously
$$\sum_{i=1}^{n}\left(\frac12i(i+1)\right)=\frac{1}{2}\left(\sum_{i=1}^{n} i^2 +\sum_{i=1}^{n}i\right).$$
Note that 
$$
\sum_{i=1}^{n} i^2=\frac{n(n+1)(2n+1)}{6}, \quad \sum_{i=1}^{n} i= \frac{n(n+1)}{2}.
$$
A: Expanding both summations is straightforward as shown in the good solutions posted. The solution below shows how the first summation is equivalent to the second without first expanding either summation. 
$$\begin{align}
\sum_{i=1}^n \frac12 i(i+1)&=\sum_{i=1}^n\sum_{j=1}^i j\\
&=\sum_{i=1}^n\sum_{j=0}^{i-1}(j+1)\\
&=\sum_{j=0}^{n-1}\sum_{i=j+1}^n(j+1)\\
&=\sum_{j=0}^{n-1}(n-j)(j+1)\\
&=\sum_{j=0}^n(n-j)(j+1)
&=\sum_{i=0}^n (n-i)(i+1)\quad\blacksquare
\end{align}$$

For the closed form solution:
$$\sum_{i=1}^n\frac 12 i(i+1)=\sum_{i=1}^n\binom {i+1}2=\binom{n+2}3\quad \blacksquare$$
