<s>[INST] 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???
 [/INST] 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 $\frac{n}2 \cdot (3+4n+8n-1)=\frac{n}2(12n+2)=n(6n+1)$
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