Simply the expression $n+(n+1)+(n+2)+(n+3)+...+(2n)$ For starters, I don't believe the geometric series sum formula can be applied? Unless I'm misunderstanding geometric series.
Anyways, I thought of splitting up each $2$-number element. i.e. consider the first element in each pair of brackets: for every pair of brackets this number is $n$. The sum goes from $n$ to $2n$, so there are $n$ number of $n$. Adding these up gives us $n^2$. 
Then the second element in each pair of brackets (note that the second element in the first pair of brackets is $0$ and in the last it's $n$, since $2n=n+n$). We have a finite series $0+1+2+3+...+ n$, whose sum is $n(n+1)/2$.
So $n+(n+1)+(n+2)+(n+3)+...+(2n)$ can be simplified to $n^2 +n(n+1)/2$. Is that correct? 
My textbook says the answer is $(n+1)n +$ $n+1 \choose 2$, which I don't understand and doesn't seem to match what I have...
 A: From $n$ to $2n$, there are $n+1$ numbers.
A: 
For starters, I don't believe the geometric series sum formula can be applied? 

Right ... but it is an arithmetic series:
$n +(n+1) + (n+2) + ... + (2n)= (0+n) + (1 + n) + (2+n) ... +(n+n) =$
$(0+1+2+3...+n)+(n+1)n = \frac{n(n-1)}{2}+(n+1)n$ 
So note that you have $n+1$ terms, rather than $n$
A: Your answer is almost right, you have that:
\begin{align}
&\;n+(n+1)+(n+2)+\dots 2n\\
=&(n+0)+(n+1)+(n+2)+\dots 2n\\
=&\;\,0+1+2+\dots+n+(n+1)n\;\;\text{ since there are }n+1\text{ numbers from } 0\text{ to } n\\
&=\frac{n(n+1)}{2}+n(n+1)
\end{align}
And as:
\begin{align}
{n+1}\choose{2}&=\frac{(n+1)!}{2!\cdot(n+1-2)!}\\
&=\frac{(n+1)\cdot n\cdot (n-1)1}{2!\cdot (n-1)1}\\
&=\frac{n(n+1)}{2!}\\
&=\frac{n(n+1)}{2}
\end{align}
You have basicially the same result as your book :)
A: You miscounted the number of elements between $n$ and $2n$, there are $n+1$ and not $n$, so instead of $n^2$ you should have $n(n+1)$. Furthermore. let's examine the n-chose-k definition:
$$
\left(\begin{array}{c}{n+1} \\ {2}\end{array}\right)=\frac{(n+1) !}{(n+1-2) ! 2 !}=\frac{1 \cdot 2 \cdot 3 \ldots(n-1) n(n+1)}{1 \cdot 2 \cdot 3 \ldots(n-1) \cdot 2}=\frac{n(n+1)}{2}
$$
So beside the little counting error your reasoning was good, you just used a different notation!
A: Using the asociativity, we have
$$
n + (n + 1) + \ldots + (2 n) = 
\underbrace{n + \ldots + n}_{\text{$n + 1$ times}} + \sum_{i = 1}^n i = n (n + 1) + \frac{n (n + 1)}{2} = \frac{3 n (n + 1)}{2} = (n + 1) n + \left(\begin{matrix} n+1 \\ 2\end{matrix}\right)\mbox{. }
$$
