# [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:

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

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

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

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

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

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

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

Factor out (k+1)

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

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

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

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

• Jun 8th 2010, 10:40 PM
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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
$\displaystyle [\frac{1}{2}k(k+1)] + (k+1)$

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

Maybe imagine it like this

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

where $\displaystyle a = \frac{1}{2}k$ and $\displaystyle b = k+1$.
• 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

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

by (k+1)