summation

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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
Danny

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} $

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

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} $

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

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