1. ## a question about induction

prove that sigma(1/r(r+1)(r+2)=0.25-1/2(n+1)(n+2)
it is the proving for k+1 i have a question about, do you put both sides equal to each other and the expand/simplify one side using the fact that you have assumed it is true for k and the show that both sides equal each other. my teacher went over this recently however i belive she got it wrong, any help would be greatly appreciated.

2. $\displaystyle \sum\limits_{r=1}^{n}{\frac{1}{r(r+1)(r+2)}}=\frac {1}{4}-\frac{1}{2(n+1)(n+2)}.$

Verify that $\displaystyle n=1$ is true & assume valid your proposition for $\displaystyle n=k.$ Hence, we must prove that $\displaystyle \sum\limits_{r=1}^{k+1}{\frac{1}{r(r+1)(r+2)}}=\fr ac{1}{4}-\frac{1}{2(k+2)(k+3)}.$

Observe that,

\displaystyle \begin{aligned} \sum\limits_{r=1}^{k+1}{\frac{1}{r(r+1)(r+2)}}&=\s um\limits_{r=1}^{k}{\frac{1}{r(r+1)(r+2)}}+\sum\li mits_{r=k+1}^{k+1}{\frac{1}{r(r+1)(r+2)}} \\ & =\frac{1}{4}-\frac{1}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)} \\ & =\frac{1}{4}-\frac{1}{(k+1)(k+2)}\left( \frac{1}{2}-\frac{1}{k+3} \right) \\ & =\frac{1}{4}-\frac{1}{(k+1)(k+2)}\cdot \frac{k+1}{2(k+3)}, \end{aligned}

which is after simplification, the result we wanted to prove. $\displaystyle \blacksquare$

3. it is this simplification which i am not sure about. must we rearrange that to the form 0.25-1/2(n+1)(n+2) or can we simply set it equal to that and cancel it down?? thanks

4. You're proving the statement by mathematical induction. You're assuming your proposition valid for $\displaystyle n=1$ and for some $\displaystyle n=k.$ The next step is to prove that for $\displaystyle n=k+1$ the equality does hold, which is I've done above.

If you want to "rearrange" to get $\displaystyle \frac{1}{4}-\frac{1}{2(n+1)(n+2)},$ you'll need no induction, but telescoping series will be helpful.