Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
 A: $$\sum_{k=1}^n(1+\ldots+k)=\sum_{k=1}^n\frac{k(k+1)}2=\frac12\left(\sum_{k=1}^nk^2+\sum_{k=1}^nk\right)$$
and now
$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$$
A: $$\begin{align}
&1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)\\
&=n\cdot 1+(n-2)\cdot 2+(n-3)\cdot 3+\cdots +1\cdot n\\
&=\sum_{r=1}^n(n+1-r)r\\
&=\sum_{r=1}^n {n+1-r\choose 1}{r\choose 1}\\
&={n+2\choose 1+2}\\
&={n+2\choose 3}\\
&=\frac16 n(n+1)(n+2)
\end{align}$$
A: 
Sorry for the horrible resolution. In any case: That's Pascal's triangle. The blue is the triangular numbers. The red is the sum of the blue (can you see why?)
Now you can use the formula for the elements of Pascal's triangle: The $n$th row and $r$th column is $\dbinom nr$. (You start counting the rows and columns from 0. The rows can be counted from the left or the right, doesn't matter.)
The answer is $\dbinom{n+2}3=\dfrac{n(n+1)(n+2)}{3!}$.
A: Hint: use also that
$$
1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6
$$
$$
1 + (1+2) + \dots + (1 +2+\dots +n) =
\frac{1(1+1)}2 + \frac{2(2+1)}2 + \dots + \frac{n(n+1)}2 
\\=\frac 12 \left[
(1^2 + 1) + (2^2 + 2 ) + \dots + (n^2 + n)
\right]
\\=\frac 12 \left[
(1^2 + 2^2 + \dots + n^2) + (1 + 2 + \dots + n)
\right]
$$
A: HINT :
It is the summation of $\sum \frac {n(n+1)}2$ from 1 to n
which is equal to $\sum (\frac {n^2}2 + \frac n2)$ from 1 to n
A: I thought about this problem differently than others so far. The problem is asking you to essentially sum up a bunch of sums. So by observation, it appears that $1$ appears $n$ times, $2$ appears $n-1$ times, $3, n-2$ times and so on, with only $1$ $n$ term. So instead, let's add up a sum from $1$ to $n$ which does this. It should be of the form $\sum_{i=1}^{n} n(n+1)=\sum_{i=1}^{n}n^2+n.$ Since sums are linear, decompose this into two sums, and apply the formulas you know for the sum of the squares and the sum of the integers.
A: Sums we know:
$\sum^n_{i=1} i = 1+2+\cdots+n=\frac{n^2+n}{2}$
$\sum^n_{i=1} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$
Your sum is $$(1+2+3+ \cdots + n) + (1 + 2 + \cdots + (n-1)) + (1 + 2 + \cdots + (n-2)) + \cdots + (1)$$ $$= \sum^n_{k=1} \sum^k_{i=1} i$$ $$= \sum^n_{k=1} \frac{k^2+k}{2} = \frac 12 (\sum^n_{k=1} k^2 + \sum^n_{k=1} k)$$
NOTE: You can reorder the terms if the are a finite number of them.  So if you're going to be taking a limit as $n \to \infty$ don't do it this way.
A: The n-th partial sum of the triangular numbers as listed in http://oeis.org/A000292 .
