Recent content by Keen94

http://www.people.vcu.edu/~rhammack/BookOfProof/ The above link directs you to a book that introduces you to logic and proof writing. Complete Chapter 1: Sets (1.1-1.7) Chapter 2: Logic from Part I. Then skip ahead to Part II

Google: The Book of Proof by Richard Hammack or type in this link http://www.people.vcu.edu/~rhammack/BookOfProof/ Go to Part III section 8. You should be able to find your way a little better.

I edited the post to clear up a few steps.

Let P(p) be the statement ##\sum_{k=1}^{n} k^p= \frac{n^{p+1}}{p+1}\ +An^{p}+Bn^{p-1}+Cn^{p-2}+...## Observe that if p=1, the statement ##\sum_{k=1}^{n} k^1= \frac{n^2}{2}\ +\frac{n}{2}## , is true. Suppose ∀n∈ℤ+, if P(1),...,P(p-1),P(p) are true, then P(p+1). Observe that if p=1,...,p the...

If you hadn't pointed it out, I would have never noticed. I'll get to it.

Would I do the adjusting to the the P(p+1) sum, so that then the P(p) sum is proved?

After adjusting the index of the sum and sending the first term of the sum to the LHS we are left with ##\sum_{k=1}^r k^{p} = \frac{(1+r)^{p+1}}{p+1}+Ar^{p}+Br^{p-1}+Cr^{p-2}+...##

How hot or cold am I? I'm guessing that the last step is to adjust the index of the the sum on the LHS by subtracting the first term of the sum from both sides.

Since P(k+1) is true whenever P(1),...,P(k) are true then P(k) is true therefore ##\sum_{k=0}^r k^{p} = \frac{(1+r)^{p+1}}{p+1}+Ar^{p}+Br^{p-1}+Cr^{p-2}+...##

I suppose it follows that ##(p+2)\sum_{k=0}^r k^{p+1} = (1+r)^{p+2}-1-\left[\sum_{k=0}^r k^0 +(p+2)\sum_{k=0}^{r} k^{1}+...+\frac{(p+2)(p+1)}{2}\ \sum_{k=0}^{r} k^{p}\right]## Then ##\sum_{k=0}^r k^{p+1} = \frac{(1+r)^{p+2}}{p+2}-\frac{1}{p+2}-\frac{1}{p+2}\ \left[\sum_{k=0}^r k^0 +...

Bump

Let P(p) be the statement ##\sum_{k=1}^n k^p = \frac{n^{p+1}}{p+1}## + polynomial of degree at most p Suppose P(1) is true, we have ##\sum_{k=1}^n k^1=\frac{n^2}{2}+\frac{n}{2}## Suppose ∀n∈ℤ+, if P(1),...,P(p-1),P(p) are all true, then P(p+1) is true. So we have, ##\sum_{k=1}^n...

In other words I just have to get that crucial first term (np+1) / (p+1) and simply reason out that the remaining terms are n's with decreasing power.

You're right. Big mistake on my part. What I meant was, am I trying to rewrite my sums so that it fits the definition of the binomial theorem? Therefore I could replace my binomial theorem sums with a polynomial.

I think I see it now. Just as (1+K)p+1-kp+1=∑kp then all the other sigmas can be replaced by a polynomial of the form (1+K)n-kn. Is that were you were going with this?