# Induction involving Sum

1. May 28, 2006

Prove that $$\sum_{k=1}^{2n} (-1)^{k}(2k+1)$$ is proportional to $$n$$, and find the constant of proportionality. So I have to prove that: $$\sum_{k=1}^{2n} (-1)^{k}(2k+1) = np$$ where $$p$$ is the constant of proportionality. So for $$n =1$$ we have $$\sum_{k=1}^{2} (-1)^{k}(2k+1) = 2$$. In this case, $$p = \frac{1}{2}$$. For $$n =s$$ we have $$\sum_{k=1}^{2s}(-1)^{k}(2k+1) = -3 + 5 + ... + (-1)^{2s}(4s+1) = ps$$ which we assume to be true. Then for $$n = s+1$$ we have $$\sum_{k=1}^{2s+2} (-1)^{k}(2k+1) = p(s+1)$$ which we want to prove from the case $$n = s$$. So $$\sum_{k=1}^{2s+2}(-1)^{k}(2k+1) = (\sum_{k=1}^{2s} (-1)^{k}(2k+1))+(-1)^{2s+1}(4s+3)$$.

Am I doing this correctly? I am stuck at this part.

Thanks

2. May 28, 2006

### Tide

You left out a term in your last equation:

$$\sum^{2s+2}a_k = \sum^{2s} a_k + a_{2s+1} + a_{2s+2}$$

3. May 28, 2006

So that means $$\sum_{k=1}^{2n+2} = (\sum_{k=1}^{2s} a_{k}) + a_{2n+1}+a_{2n+2} = ps + (-1)^{2n+1}(4n+3) + (-1)^{2s+2}(4s+5) = ps-4n-3 + 4s + 5 = ps + 2$$. So $$ps+2 = ps + p$$ so does that mean $$p = 2$$?

Thanks

4. May 28, 2006

### VietDao29

Yes, p = 2, not 1 / 2. This part is wrong.
$$\sum_{k=1}^{2} (-1)^{k}(2k+1) = 2 = np = 1 \times p = p$$
So p = 2.
:)