# summation

• January 20th 2009, 01:45 PM
qzno
summation
Justify:

$
\sum_{i=n}^{2n} i = \frac{3n(n+1)}{2}
$
• January 20th 2009, 01:57 PM
Plato
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
$
• January 20th 2009, 01:58 PM
Jester
Quote:

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}
$

• January 20th 2009, 02:01 PM
Krizalid
$\sum\limits_{i=n}^{2n}{i}=\sum\limits_{i=0}^{n}{(i +n)}.$
• January 20th 2009, 02:25 PM
qzno
Quote:

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 : (
• January 20th 2009, 02:33 PM
Plato
$\sum\limits_{k = 1}^{2n} k = \sum\limits_{k = 1}^{n - 1} k + \sum\limits_{k = n}^{2n} k$

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

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

Now try it again.
• January 20th 2009, 03:20 PM
Jester
Quote:

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

Quote:

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}
$

We need the extra part - adding $n$ :)