Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$? 
Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.

The only reduction I thought about so far would be:
$$n\cdot 1 + (n - 1)\cdot 2 + (n - 3)\cdot 3 +\cdots $$
which seems symmetrical; for example, an odd and even n
 5 4 3 2 1        4 3 2 1
*                *
 1 2 3 4 5        1 2 3 4

but I'm not sure if and how it may help derive a formula.
 A: $$f(n)=\sum_{i=1}^n \frac{i(i+1)}{2}$$
$$f(n)=\sum_{i=1}^n \frac{i^2}2 +\frac{i}2$$
Using two well known identitites,
$$f(n)=\frac12\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)$$
Simplifying:
$$f(n)=\frac{n(n+1)}{4}\left(\frac{2n+1}{3}+1\right)$$
$$f(n)=\frac{n(n+1)}{4}\left(\frac{2n+4}{3}\right)$$
$$f(n)=\frac{n(n+1)}{2}\left(\frac{n+2}{3}\right)$$
$$f(n)=\frac{n(n+1)(n+2)}{6}$$
A: There is a general formula $$\sum_{k=1}^n\frac {k(k+1)\dots(k+r-1)}{r!}=\frac {n(n+1)\dots(n+r)}{(r+1)!}$$
Which can be proved by induction - base case $n=1$, both sides of the equation are equal to $1$.
Then $$\frac {n\left[(n+1)\dots(n+r)\right]}{(r+1)!}+\frac {\left[(n+1)\dots(n+r)\right]}{r!}=\frac {\left[(n+1)\dots(n+r)\right](n+r+1)}{(r+1)!}$$
Another way of writing this, which invites a combinatorial proof is:
$$\sum_{k=1}^n\binom {k+r-1}r=\binom {n+r}{r+1}$$
The sum of integers is the case $r=1$ and the sum of triangular numbers is the case $r=2$.
A: I think the initial steps towards a formula look a bit simpler than seen so far here.
Beginning with
$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$
we count the number of occurences of the $1$, of the $2$
$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$
and rewrite a bit
$$ \begin{align} H^{(2)}(n) &= 1\cdot ((n+1)-1) + 2 \cdot ((n+1)-2) + 3\cdot ((n+1)-3) + ... + n \cdot ((n+1)-n) \\ \qquad \\
&= (1+2+3+...+n)(n+1) - (1^2+2^2+3^2+...+n^2) \\ \qquad \\
&= {n(n+1) \over 2}(n+1) - {2n^3 + 3n^2 + 1n\over 6} \end{align}$$
The last steps stems from the faulhaber formulae for sums-of-like-powers.

Final step, expanding & collecting this gives the same result as derived by other answerers:
$$ H^{(2)}(n) = {3n(n+1)(n+1)-2n^3 - 3n^2 - 1n \over 6} \\
= {n^3+3n^2+2n \over 6} \\
= {n(n+1)(n+2) \over 3!}  $$
For $n=5$ we get the sum $H^{(2)}(5)=35$
A: The hockey stick identity states that
$$\sum_{k=r}^{n} \binom{k}{r}=\binom{n+1}{r+1}$$
Note that $$1+2+...+n = \frac{n(n+1)}{2}=\binom{n+1}{2}.$$
Hence
$$f(n)=\binom{2}{2}+\binom{3}{2}+...+\binom{n+1}{2}=\binom{n+2}{3}=\frac{(n+2)(n+1)n}{3!}$$
