Help understanding $1+2+3+\dots+n$, proof I am relatively new to maths, and I have hard time understanding and visualizing part of this particular proof for the sum of $1+2+3+\dots+n$.
If we take the 1st number and the Nth number and add them together we should get $n/2$ such pairs, arriving at the formula $(n/2)(n+1)$.
My question is, how do I understand this for odd numbers since there is nothing to pair it with.
Take for example $1$ to $100$, there are $50$ such pairs each amounting to $101$.
But for odd numbers,
Take for example $1$ to $101$, the number $51$ has no such pairing, what is the significance of the middle number?
So for odd numbers in this case, the middle number is always equal to $(n+1)/2$, why is this so?
Is there a way of understanding this or visualizing this? Thanks.
edit: made a mistake, middle number, not necessarily odd. But the Nth number is odd in this case.
 A: It’s a little easier to visualize if you imagine writing out the sum twice, in opposite orders, like this:
$$\begin{array}{ccc}
S&=&1&+&2&+&\ldots&+&n-1&+&n\\
S&=&n&+&n-1&+&\ldots&+&2&+&1\\\hline
2S&=&(n+1)&+&(n+1)&+&\ldots&+&(n+1)&+&(n+1)
\end{array}$$
The righthand side of the bottom line is $n(n+1)$: there are $n$ columns, and each column contains a pair of numbers that sum to $n+1$. Thus, $2S=n(n+1)$, and therefore
$$S=\frac{n(n+1)}2\;.$$
When $n$ is even you can simply take half of the columns: then you have $\frac{n}2$ columns, each summing to $n+1$, and every integer $k$ from $1$ through $n$ has appeared exactly once, paired with $(n+1)-k$, so you get the same formula without having to solve for $S$. The remaining $\frac{n}2$ columns in the array above contain the same pairs of numbers, but in the opposite order, with the larger one on top. Thus, when $n$ is even it’s reasonable to think of pairing each $k$ with $(n+1)-k$.
But as you’ve seen, when $n$ is odd that doesn’t work, because the middle number is its own ‘mate’ in the pairing of $k$ with $(n+1)-k$: when $k=\frac{n+1}2$, $(n+1)-k$ is also $\frac{n+1}2$. It’s easier to double up, as I did above, and count every integer in the progression twice. And when we write the two copies in opposite orders, every number, including the one in the middle when $n$ is odd, automatically gets paired with its proper mate.
To see why the middle number is $\frac{n+1}2$, let $n=2m-1$. (Remember, $n$ is odd, so we can certainly write it this way.) There are $m-1$ integers from $1$ through $m-1$, and there are also $m-1$ integers from $m+1$ through $m+(m-1)=2m-1=n$:
$$\underbrace{\color{red}{1,2,3,\ldots,m-1}}_{m-1\text{ integers}},m,\underbrace{\color{blue}{m+1,m+2,m+3,\ldots,m+(m-1)}}_{m-1\text{ integers}}$$
That accounts for $2(m-1)=2m-2=n-1$ of the numbers from $1$ through $n$, and the integer in the middle is $m=\frac{n+1}2$.
A: You are right, if you have an odd number $n$ there is nothing that you can  pair with the $\frac{n+1}2$-th number. The average $\frac{n+1}2$ is the middle number because its distance to $n$ is
$$n-\frac{n+1}2=\frac{n-1}2,$$ and this is the same as its distance to $1$, which is $$\frac{n+1}2-1=\frac{n-1}2.$$ So you have $\frac{n-1}2$-pairs that sum up to $n+1$ and the number $\frac{n+1}2$. So the sum of all these numbers is
$$\frac{n-1}2 (n+1) + \frac{n+1}2=\frac{n(n+1)}2.$$
It is much simpler to calculate
$$\begin{array}{r}S_n&=&1&+&2&+&3&+&\ldots&+&(n-2)&+&(n-1)&+&n \\
S_n&=&n&+&(n-1)&+&(n-2)&+&\ldots&+&3&+&2&+&1\\
\hline\\
2S_n&=&(n+1)&+&(n+1)&+&(n+1)&+&\ldots&+&(n+1)&+&(n+1)&+&(n+1)
\end{array}$$
and so
$$2S_n=(n+1)n$$
and further
$$S_n=\frac{(n+1)n}2$$
Here a proof without words from https://www.maa.org/sites/default/files/Richards31975.pdf

The following picture from http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html is the visulaization om my proof:

A: 
So $2 \cdot S = n \cdot (n+1) \Rightarrow \boxed{S = \frac{n \cdot (n+1)}{2}}$
A: Let's take this series for example  1+2+3+4+5+6+7+8+9+10 =55
and n be no. of terms. Now add first term and last term, second term and second last term and so on you will observe that their sum will be equal to 11 so using some logic since you have made 5 pairs  you will multiply 11 by 5 giving 55 hence for even number of terms the sum is (n/2)(first term + last term).
Now let's take this series 1+2+3+4+5+6+7+8+9=45
Again add first term and last term, second term and second last term and so on you will observe that their sum will be equal to 10 and 5 will be left out so 4 pairs are formed thus we have {(n-1)/2}(first term + last term )+(n+1)/2 and by some basic algebra we get it equal to (n/2)(first term+ last term)
A: I think a familiar object that may help see this sum is a seesaw.
The middle point, for $n$ odd, is the pivot point. For $n$ even, the midpoint would be between the paired points $\frac{n}{2}$ and $\frac{n}{2}+1$.
The points that are paired are at the same distance from the pivot point.
When we write $1+2+...+n$, the seesaw is in one extreme position, say the kid on the left is down and the kid on the right, up.
When we write $n+(n-1)+...+1$, the seesaw is in the opposite position: left kid up, right kid down.
If both kids get out of the seesaw and somehow we set the bar horizontally, the height of every point is set to the average height, or height of the midpoint, $\frac{n+1}{2}$. And there are $n$ such points on the seesaw, hence the result.
We can think of the middle point for $n$ odd as the only point that already has the average height with no need for compensation from the other side.
A: There are $50.5$ such pairs, each summing to $102$.
A: Managed to figure out the answer
Draw it in increasing square units and chop it in half, it cuts neat for even, but odd cuts the (middle) squares in half! so when you match em you get N+1(height)/2.
S.D.G.
