# Thread: Mathmatical induction

1. ## Mathmatical induction

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!

2. ## Re: Mathmatical induction

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?

3. ## Re: Mathmatical induction

Bad bracketing...repost clearly...

4. ## Re: Mathmatical induction

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...

5. ## Re: Mathmatical induction

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!

6. ## Re: Mathmatical induction

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 $\frac{1}{2} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$.

7. ## Re: Mathmatical induction

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))