Why does $ 1+2+3+\cdots+p = {(1⁄2)}\cdots(p+1) $ I saw this from Project Euler, problem #1:
If we now also note that $ 1+2+3+\cdots+p = {(1/2)} \cdot p\cdot(p+1) $
What is the intuitive explanation for this?  How would I go about deriving the latter from the former?  It has the summation express which I am not sure how to deal with, so unfortunately I am not even sure how I would begin to show these are equivalent.
 A: $\begin{array}{cccccc}
1 & 2 & \cdots & p-1 & p\\
p & p-1 &\cdots  & 2 & 1\\
-- & -- & -- & -- & -- & +\\
p+1 & p+1 &\cdots  & p+1 & p+1\end{array}$
Counting twice you get: $p\times\left(p+1\right)$
So counting once you get:  $\frac{1}{2}\times p\times\left(p+1\right)$ 
A: Gauss first came up with a sleek way to do it: he grouped the $1$st term and the last term:
$1 + p$, and then the $2$nd term $2$ and the second to last term $p-1$, and so on until he 
reached the middle terms. He found that all the groups have the same sum: $p+1$, and there 
were a total of $\dfrac{p}{2}$ groups. So the answer is what you got.
A: There’s a nice way to see this which allegedly comes from Gauss: his teacher asked the class to sum the numbers from $1$ to $100$, and Gauss found a neat trick.
Write $S_p$ for the sum $1 + 2 + \cdots + p$. If we write it out once, and then again in reverse:
$$
\begin{align*}
S_p &= 1 + \;\;\;\,2\;\;\;\; + \cdots + p \\
S_p &= p + (p-1) + \cdots + 1
\end{align*}
$$
then we see there’s a nice way to pair up terms:
$$S_p + S_p = 2S_p = (p+1) + (p+1) + \cdots + (p+1) = p\,(p+1)$$
We can then divide by $2$ to get the result:
$$S_p = \tfrac{1}{2}p\,(p+1)$$
A: Let       S=1 +2   +3    +...(p-1)+p
also,     S=p+(p-1)+(p-2)+...2    +1
  we know that the total number of terms in S =p 
  because these are are first p natural numbers 
  See my spacing between + signs in above two expression
   now add them 
   you will get
  2S=(p+1)+(p+1)+(p+1)+...(p+1)   upto p number of terms
   also multiplication is repetetive addition
   so 2S=p(p+1)
implies S=1/2(p(p+1))= S=1+2+3+...(p-1)+p
A: Proof without words in Wolfram Demonstrations Project:
http://demonstrations.wolfram.com/ProofWithoutWords12N1NChoose2/.
A: Alternatively, you can think of it like this
.
. .
. . .
. . . .
<etc...>
. . . . . . . . <etc...> . (n dots here)

For example, when $n = 6$, we have
. 
. .
. . .
. . . .
. . . . .
. . . . . .

Now, note that the ith row contains i dots, so the total number of dots is equal to $1 + 2 + ... + n$. (in this case, n = 6) Now, the area of the right triangle is $\frac{n(n + 1)}{2}$. This gives us the answer.
Alternatively, you can prove this by induction.
