Sum of Arithmetic progression problem

I have following progression:

$2n+(2n-1)+\cdots+n$

which is equivalent to:

$n+(n+1)+\cdots+2n$

You match $2n-i$ with $n+i$ for all $i=0..n$ Therefore you have $(n+1)$ terms of $3n$, so the sum of the $2$ exactly same sequence is $3n(n+1)$ and therefore the sum of 1 sequence is $1.5n(n+1)$

Or you can just apply the formula for the sum of Arithmetic progression, please refer to the wiki page https://en.wikipedia.org/wiki/Arithmetic_progression

Thus, I tried applying sum of arithmetic progression since I need to make it "$1.5n(n+1)$". but when I apply sum of arithmetic progression, it gives me different result like "$\frac n2(2n+(n-1)*1) = 1.5n^2-0.5n$"

How can I get "$1.5n(n+1)$"?

• Welcome to MSE! Please Use Mathjax Commented Jun 15, 2017 at 12:35
• To correctly display formulas enclose them with dollar signs \$\$ see my edit to see how,for a bit more advanced tutorial for typing math on this site look at this tutorial MathJax basic tutorial and quick reference Commented Jun 15, 2017 at 12:53

Applying the formula, $Sum=\frac{(n+2n)*(n+1)}{2}$ reaches your desired result...