The proof behind this pattern

arithmeticpalindrome

Let me get to the point.
We see that 121 is a palindrome, also 12321 is a palindrome.
If we were to add up the digits in each palindrome, for example, 121 we get 4. If we square root four, the root is the middle number of 121. This pattern goes on from what I worked on two years ago. Another example:
1234321 <= Palindrome
1+2+3+4+3+2+1 = 16 <= Palindrome's Digits Sum
square root of 16 = 4
Also the number 11 is excluded from this pattern for obvious reason, is there any logic behind this exclusion?
What is the real mathematics behind this pattern?
EDIT:
The palindromes are digits ranging from one to n and the same n to one. (1 + n1 + n2 + n + n2 + n1 + 1)
121
12321
1234321
123454321
This is what is meant by palindromes.
EDIT:
I need to clarify based on the comments. When I heard about palindromes and ran into this pattern I associated these digit sequences as palindromes. These numbers are symmetrical in terms that there is a middle digit and the digits left and right are descending being a digit till one is reached( 1, 2, 3, 2, 1). Sorry if I have mistaken these numbers as palindromes.

Best Answer

Sum from $1$ to $n$ is given by

$$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$

Sum from $1$ to $n$ and then $n$ to $1$ is

$$ 1 + 2 + \cdots + n + (n-1) + \cdots + 1 = \frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2 + n + n^2 - n}{2} = \frac{2n^2}{2} = n^2.$$

Take the square root and you get $n$, your middle number.

Reference: Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$

Related Question