I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$ I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$
$\sum_{i=n}^{2n-1} 3+  \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ am I right?

$4 \sum_{i=0}^{2n-1-n} i+n $ 
$4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )$ = $4(\frac{n(n-1)}{2}+n(n-1))$=$2(3n^2-3n)=6n^2-6n$
total= $3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $=$3n+ 6n^2-6n $, am I right???
 A: There are $\color{#C00}{n}$ terms and the average of $i$ is $\color{#090}{\frac{3n-1}2}$
$$
\begin{align}
\sum_{i=n}^{2n-1}(3+4i)
&=3\color{#C00}{n}+4\color{#C00}{n}\color{#090}{\frac{3n-1}2}\\
&=n(6n+1)
\end{align}
$$
A: Alternatively:
$$\begin{align}\sum_{i=n}^{2n-1}(3+4i)&=\sum_{i=0}^{2n-1}(3+4i)-\sum_{i=0}^{n-1}(3+4i)=\\
&=3\cdot 2n+4\cdot \frac{2n(2n-1)}{2}-3n-4\cdot \frac{n(n-1)}{2}= \\
&=6n+8n^2-4n-3n-2n^2+2n=\\
&=6n^2+n=\\
&=n(6n+1).\end{align}$$
A: Note $4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )=4 \left( \frac{n(n-1)}{2}+\color{red}{n^2}\right)$ since from $0$ to $n-1$ there are $n$ numbers.
Another comment is use braces 
$4 \sum_{i=0}^{2n-1-n} i+n $  should be $4 \sum_{i=0}^{2n-1-n} (i+n) $ 
Alternative approach: Note that you are summing an arithmetic progression.
There are a total of $(2n-1-n+1)=n$ terms.
The first term is $3+4n$, the last term is $3+4(2n-1)=8n-1$.
Hence the sum is $\frac{n}2 \cdot (3+4n+8n-1)=\frac{n}2(12n+2)=n(6n+1)$
A: That is how you do it, yes.
I suppose that as I am the type of person who always makes an indexing area to play it safe I'd to this
$\sum_{i=n}^{2n-1} 3+4i = $
$\sum_{j= 0}^{n-1} (3 + 4(j + n)) = $
$3\sum_{j= 0}^{n-1} 1 + 4\sum_{j= 0}^{n-1} j + n\sum_{j= 0}^{n-1} 1=$
$3*n + 4(\frac {n(n-1)}2) + n*n = $
$3n + 2n(n-1) + n^2 = $
$3n + 2n^2 - 2n + n^2 = $
$3n^2 + n$
....
You have an error at 
$4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )=4(\frac{n(n-1)}{2}+n*n)$
A: Pedestrian approach:
$\sum_{i=n}^{2n-1}i =$   
$n \ \ \ \ + \ \ \ \ \  \ \  \ (n+1) +.....(2n-1)$.
$(2n-1) + (2n-2)+.......n =$
$\sum_{i=1}^{2n-1}i$  (in reverse order).
Adding the sums:
$2\sum_{i=n}^{2n-1}i = (n)(3n-1)$.
Thus:
$\sum_{i=n}^{i=2n-1}(3+4i)=$
$3(n)+4(n/2)(3n-1)=$
$ 3n +2n(3n-1)=6n^2+n$.
A: 
Hint: Note the brackets $\sum_{i=n}^{2n-1}\color{blue}{(} 3+4i\color{blue}{)}$  @robjohn used in his answer. It is crucial to use them, since the sum without brackets is 
  \begin{align*}
\color{blue}{\sum_{i=n}^{2n-1} 3+4i}
=\left(\sum_{i=n}^{2n-1} 3\right)+4i
=3\left(\sum_{i=n}^{2n-1} 1\right)+4i\color{blue}{=3n+4i}
\end{align*}
  with $i$ either the imaginary unit or with $i$ a variable having the same scope as $n$.

