# [SOLVED] algebra in P.M.I

• Jun 8th 2010, 09:22 PM
dunsta
[SOLVED] algebra in P.M.I
I am still having trouble following the simplification of equations in the Induction examples i am studying.
The notes I have use this:

$
[\frac{1}{2}k(k+1)] + (k+1)
$

which is simplified to
$
\frac{1}{2}(k+1)[(k+1)+1]
$

but when I do the math I get
$
[\frac{1}{2}k(k+1)] + (k+1) = [\frac{1}{2}k^2 + k] + (k+1) = [\frac{1}{2}k^2 + k] + (k+1)
$

so if anyone could explain the steps used in the former expression I would me much appreciative
• Jun 8th 2010, 10:10 PM
undefined
Quote:

Originally Posted by dunsta
I am still having trouble following the simplification of equations in the Induction examples i am studying.
The notes I have use this:

$
[\frac{1}{2}k(k+1)] + (k+1)
$

which is simplified to
$
\frac{1}{2}(k+1)[(k+1)+1]
$

but when I do the math I get
$
[\frac{1}{2}k(k+1)] + (k+1) = [\frac{1}{2}k^2 + k] + (k+1) = [\frac{1}{2}k^2 + k] + (k+1)
$

so if anyone could explain the steps used in the former expression I would me much appreciative

$[\frac{1}{2}k(k+1)] + (k+1)$

Factor out (k+1)

$(k+1)(1+\frac{1}{2}k)$

You can get a 1/2 on the outside by multiplying by 1 = (1/2)(2)

$\frac{1}{2}(k+1)(2+k)$

$
\frac{1}{2}(k+1)[(k+1)+1]
$
• Jun 8th 2010, 10:35 PM
dunsta
I am getting there slowly, thanks. I am sorry for my ignorance.

I know what factoring is, but could you explain the first step of how to factor out
$[\frac{1}{2}k(k+1)] + (k+1)
$

• Jun 8th 2010, 10:40 PM
undefined
Quote:

Originally Posted by dunsta
I am getting there slowly, thanks. I am sorry for my ignorance.

I know what factoring is, but could you explain the first step of how to factor out
$[\frac{1}{2}k(k+1)] + (k+1)
$

Sure. We have two terms (addends) that are both divisible by (k+1).

Maybe imagine it like this

$ab + b = b(a + 1)$

where $a = \frac{1}{2}k$ and $b = k+1$.
• Jun 9th 2010, 05:18 AM
dunsta
Ok, got it, just needed to refresh my brain of some basic algebra!!

Thanks, I understand the expression, but I still can't see how to divide

$[\frac{1}{2}k(k+1)] + (k+1)
$

by (k+1)