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!
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?
Why is this in Geometry when it's clearly Algebra? Anyway, it doesn't...
$\displaystyle \displaystyle \begin{align*} \frac{1}{2} - \frac{1}{k(k + 1)} + \frac{2}{(k + 1)(k + 2)} &= \frac{1}{2} - \frac{k + 2}{k( k + 1)(k + 2)} + \frac{2k}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{2k - k - 2}{k(k + 1)(k + 2)} \\ &= \frac{1}{2} + \frac{k - 2}{k(k + 1)(k + 2)} \end{align*}$
which does NOT equal $\displaystyle \displaystyle \begin{align*} \frac{1}{2} - \frac{1}{(k + 1)(k + 2)} \end{align*}$
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...
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!
This is a good lessen: ALWAYS, ALWAYS post the entire exact question.
By not doing so, you have a wasted much of time.
$\displaystyle \sum\limits_{r = 2}^n {\frac{2}{{r\left( {{r^2} - 1} \right)}}} = \frac{1}{2} - \frac{1}{{n\left( {n + 1} \right)}}$
Now you want to go from $\displaystyle \frac{1}{2} - \frac{1}{{n\left( {n + 1} \right)}} + \frac{2}{{\left( {n + 1} \right)\left( {{{\left( {n + 1} \right)}^2} - 1} \right)}}$ to $\displaystyle \frac{1}{2} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$.