1. ## summation

Justify:

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

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

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

so subtracting the first from the second and adding n gives your answer. (Too slow :-))

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

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

If you subtracted it wouldn't you get:

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

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

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

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

That's not what it says : (

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

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

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

Now try it again.

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

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

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

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

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

That's not what it says : (
yes but

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

so subtracting the first from the second and adding n gives your answer. (Too slow :-))
We need the extra part - adding $n$