I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$

Question

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???

Answer 1

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

Answer 2

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 $\frac{n}2 \cdot (3+4n+8n-1)=\frac{n}2(12n+2)=n(6n+1)$

Answer 3

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

$4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )=4(\frac{n(n-1)}{2}+n*n)$

Answer 4

There are $\color{#C00}{n}$ terms and the average of $i$ is $\color{#090}{\frac{3n-1}2}$ $$ \begin{align} \sum_{i=n}^{2n-1}(3+4i) &=3\color{#C00}{n}+4\color{#C00}{n}\color{#090}{\frac{3n-1}2}\\ &=n(6n+1) \end{align} $$

Answer 5

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$.

Answer 6

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 $n$.