1. ## summation

Justify:

$\displaystyle \sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}$

2. Recall the $\displaystyle \sum\limits_{k = 1}^J k = \frac{{J\left( {J + 1} \right)}}{2}$.
Then note that $\displaystyle \sum\limits_{k = 1}^{2n} k = \sum\limits_{k = 1}^{n - 1} k + \sum\limits_{k = n}^{2n} k$

3. Originally Posted by qzno
Justify:

$\displaystyle \sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}$
Since
$\displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}$

4. $\displaystyle \sum\limits_{i=n}^{2n}{i}=\sum\limits_{i=0}^{n}{(i +n)}.$

5. Originally Posted by danny arrigo
Since
$\displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}$

If you subtracted it wouldn't you get:

$\displaystyle \frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}$

$\displaystyle \frac{4n^2 + 2n - n^2 - n}{2}$

$\displaystyle \frac{3n^2 + n}{2}$

$\displaystyle \frac{n(3n+1)}{2}$

That's not what it says : (

6. $\displaystyle \sum\limits_{k = 1}^{2n} k = \sum\limits_{k = 1}^{n - 1} k + \sum\limits_{k = n}^{2n} k$

$\displaystyle \sum\limits_{k = 1}^{2n} k = \frac{{2n\left( {2n + 1} \right)}} {2}$

$\displaystyle \sum\limits_{k = 1}^{n - 1} k = \frac{{\left( {n - 1} \right)n}} {2}$

Now try it again.

7. Originally Posted by qzno
If you subtracted it wouldn't you get:

$\displaystyle \frac{2n(2n+1)}{2} - \frac{n(n+1)}{2}$

$\displaystyle \frac{4n^2 + 2n - n^2 - n}{2}$

$\displaystyle \frac{3n^2 + n}{2}$

$\displaystyle \frac{n(3n+1)}{2}$

That's not what it says : (
yes but

Originally Posted by danny arrigo
Since
$\displaystyle \sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^{2n} i = \frac{2n(2n+1)}{2}$

We need the extra part - adding $\displaystyle n$