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

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

Thanks

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# Homework Help: Induction involving Sum

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