Mathmatical induction

**gurnster76** · May 25th 2012, 01:00 AM

how do i get from

(1/2) - (1/k(k+1)) + (2/(k+1)(k+2)

to (1/2) - (1/(k+1)(k+2)

Any help would be great!

**emakarov** · May 25th 2012, 01:18 AM

What do you mean by "get from ... to ..."? The equality of these two expressions seems to hold only for k = 1: WolframAlpha.

Could you post the original problem?

**Wilmer** · May 25th 2012, 01:21 AM

Bad bracketing...repost clearly...

**Prove It** · May 25th 2012, 01:22 AM

Originally Posted by gurnster76

how do i get from

(1/2) - (1/k(k+1)) + (2/(k+1)(k+2)

to (1/2) - (1/(k+1)(k+2)

Any help would be great!

Why is this in Geometry when it's clearly Algebra? Anyway, it doesn't...

$\displaystyle \begin{align*} \frac{1}{2} - \frac{1}{k(k + 1)} + \frac{2}{(k + 1)(k + 2)} &= \frac{1}{2} - \frac{k + 2}{k( k + 1)(k + 2)} + \frac{2k}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{2k - k - 2}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{k - 2}{k(k + 1)(k + 2)} \end{align*}$

which does NOT equal $\displaystyle \begin{align*} \frac{1}{2} - \frac{1}{(k + 1)(k + 2)} \end{align*}$

I expect you might have made a mistake somewhere in your induction process, which is why it's advisable to always post the ENTIRE question and ALL of your working...

**gurnster76** · May 25th 2012, 05:02 AM

sorry for the bad bracketing, this is the original question, im still massively confused by this.

Use mathematical induction to prove that

n
Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
r=2

i have done the basic proof and got to the point

k+1
Σ = (1/2) - (1/k(k+1)) + 2/(k+1)(k+1^2-1)
r=2

I think i need to get to (1/2) - 1/(k+1)(k+2) but im not sure how to get there, any help would be greatly appreciated!

**Plato** · May 25th 2012, 05:51 AM

Originally Posted by gurnster76

sorry for the bad bracketing, this is the original question, im still massively confused by this.
Use mathematical induction to prove that
n
Σ 2/r(r^2-1) = (1/2) - (1/n(n+1)) n= 2,3,4.......
r=2

This is a good lessen: ALWAYS, ALWAYS post the entire exact question.
By not doing so, you have a wasted much of time.

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

Now you want to go from $\frac{1}{2} - \frac{1}{{n\left( {n + 1} \right)}} + \frac{2}{{\left( {n + 1} \right)\left( {{{\left( {n + 1} \right)}^2} - 1} \right)}}$ to $\frac{1}{2} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$ .

**Wilmer** · May 25th 2012, 07:21 AM

Originally Posted by gurnster76

2/r(r^2-1)

That means 2 divided by r, then the result multiplied by (r^2-1)

Do you mean 2 divided by the whole thing; if so, this is proper way: 2/(r(r^2-1))

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