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???

Alternatively:
$$\begin{align}\sum_{i=n}^{2n-1}(3+4i)&=\sum_{i=0}^{2n-1}(3+4i)-\sum_{i=0}^{n-1}(3+4i)=\\
&=3\cdot 2n+4\cdot \frac{2n(2n-1)}{2}-3n-4\cdot \frac{n(n-1)}{2}= \\
&=6n+8n^2-4n-3n-2n^2+2n=\\
&=6n^2+n=\\
&=n(6n+1).\end{align}$$
