1. ## Proof by Induction

Use the method of mathematical induction to prove that, for $n \in Z$

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

I've done the first stage, which is to show that it's true for n=1 (both sides are equal to 2 in said situation). I can easily do the assumption stage, which is simply to substitute in n=k. I know that the next stage involves showing it's true for n=k+1 but I can't seem to do this correctly. I think my memory is slightly fuzzy and my notes are just confusing me. Any help would be appreciated.

Edit: woops, posted this in pre-calc. >.< move this to pre-algebra and algebra if necessary, but please don't delete the thread - figuring out that latex took me hours.

2. So if I remember right the next step would be:

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

I can't seem to get it to work out, it's been a while, but you want to re-arrange the right side to look like:

$\frac{1}{2}((n+1)^{2} + (n+1) +4) - (\frac{1}{2})^{n}$

Sorry I couldn't be more help .

3. P(k)

$\displaystyle\sum_{r=1}^k\left(r+\frac{1}{2}^{r-1}\right)=\frac{1}{2}\left(k^2+k+4\right)-\frac{1}{2}^{k-1}$

P(k+1)

$\displaystyle\sum_{r=1}^{k+1}\left(r+\frac{1}{2}^{ r-1}\right)=\frac{1}{2}\left[(k+1)^2+(k+1)+4\right]-\frac{1}{2}^{k}$

Try to show that P(k) if true will cause P(k+1) to also be true.

Proof

$\displaystyle\sum_{r=1}^{k+1}\left(r+\frac{1}{2}^{ r-1}\right)=\left[\frac{1}{2}\left(k^2+k+4\right)-\frac{1}{2}^{k-1}\right]+(k+1)+\frac{1}{2}^k$

Now try to write the RHS of P(k+1) in terms of this to complete the inductive step, by showing P(k+1) to be true

4. Thanks for your time, I now have the solution.