Proving the Proportionality of a Sum Involving Induction

[courtrigrad]

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

 

[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}

 

[courtrigrad]

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

 

[VietDao29]

Yes, p = 2, not 1 / 2. This part is wrong.

So for n =1 we have \sum_{k=1}^{2} (-1)^{k}(2k+1) = 2. In this case, p = \frac{1}{2}

\sum_{k=1}^{2} (-1)^{k}(2k+1) = 2 = np = 1 \times p = p
So p = 2.
:)