Question Number 145664 by qaz last updated on 07/Jul/21
$$\mathrm{S}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{k}^{\mathrm{3}} =? \\ $$
Answered by mathmax by abdo last updated on 07/Jul/21
![S=Σ_(keven) (...)+Σ_(k odd) (...) =Σ_(p0) ^([(n/2)]) (2p)^3 −Σ_(p=0) ^([((n−1)/2)]) (2p+1)^3 =8Σ_(p=0) ^([(n/2)]) p^(3 ) −Σ_(p=0) ^([((n−1)/2)]) (8p^3 +3(2p)^2 +3(2p)+1) =8Σ_(p=0) ^([(n/2)]) p^3 −8Σ_(p=0) ^([((n−1)/2)]) p^3 −12Σ_(p=0) ^([((n−1)/2)]) p^2 −6Σ_(p=0) ^([((n−1)/2)]) p−([((n−1)/2)]+1) let α(n)=[(n/2)] and β(n)=[((n−1)/2)] ⇒ S=8×((α^2 (n)(α(n)+1)^2 )/4)−8×((β_n ^2 (β(n)+1)^2 )/4)−12×((β(n)(β(n)+1)(2β(n)+1))/6) −6×((β(n)(β(n)+1))/2)−β(n)−1 ⇒S=2α_n ^2 (α(n)+1)^2 −2β_n ^2 (β(n)+1)^2 −2β(n)(β_n +1)(2β_n +1) −3β(n)(β(n)+1)−β_n −1](https://www.tinkutara.com/question/Q145709.png)
$$\mathrm{S}=\sum_{\mathrm{keven}} \left(…\right)+\sum_{\mathrm{k}\:\mathrm{odd}} \left(…\right) \\ $$$$=\sum_{\mathrm{p0}} ^{\left[\frac{\mathrm{n}}{\mathrm{2}}\right]} \left(\mathrm{2p}\right)^{\mathrm{3}} \:−\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \left(\mathrm{2p}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$=\mathrm{8}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}}{\mathrm{2}}\right]} \:\mathrm{p}^{\mathrm{3}\:\:} −\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\left(\mathrm{8p}^{\mathrm{3}} \:+\mathrm{3}\left(\mathrm{2p}\right)^{\mathrm{2}} \:+\mathrm{3}\left(\mathrm{2p}\right)+\mathrm{1}\right) \\ $$$$=\mathrm{8}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}}{\mathrm{2}}\right]} \:\mathrm{p}^{\mathrm{3}} −\mathrm{8}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \mathrm{p}^{\mathrm{3}} −\mathrm{12}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{p}^{\mathrm{2}} −\mathrm{6}\sum_{\mathrm{p}=\mathrm{0}} ^{\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{p}−\left(\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}\right) \\ $$$$\mathrm{let}\:\alpha\left(\mathrm{n}\right)=\left[\frac{\mathrm{n}}{\mathrm{2}}\right]\:\mathrm{and}\:\beta\left(\mathrm{n}\right)=\left[\frac{\mathrm{n}−\mathrm{1}}{\mathrm{2}}\right]\:\Rightarrow \\ $$$$\mathrm{S}=\mathrm{8}×\frac{\alpha^{\mathrm{2}} \left(\mathrm{n}\right)\left(\alpha\left(\mathrm{n}\right)+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{8}×\frac{\beta_{\mathrm{n}} ^{\mathrm{2}} \left(\beta\left(\mathrm{n}\right)+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}−\mathrm{12}×\frac{\beta\left(\mathrm{n}\right)\left(\beta\left(\mathrm{n}\right)+\mathrm{1}\right)\left(\mathrm{2}\beta\left(\mathrm{n}\right)+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$−\mathrm{6}×\frac{\beta\left(\mathrm{n}\right)\left(\beta\left(\mathrm{n}\right)+\mathrm{1}\right)}{\mathrm{2}}−\beta\left(\mathrm{n}\right)−\mathrm{1} \\ $$$$\Rightarrow\mathrm{S}=\mathrm{2}\alpha_{\mathrm{n}} ^{\mathrm{2}} \left(\alpha\left(\mathrm{n}\right)+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\beta_{\mathrm{n}} ^{\mathrm{2}} \left(\beta\left(\mathrm{n}\right)+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\beta\left(\mathrm{n}\right)\left(\beta_{\mathrm{n}} +\mathrm{1}\right)\left(\mathrm{2}\beta_{\mathrm{n}} +\mathrm{1}\right) \\ $$$$−\mathrm{3}\beta\left(\mathrm{n}\right)\left(\beta\left(\mathrm{n}\right)+\mathrm{1}\right)−\beta_{\mathrm{n}} −\mathrm{1} \\ $$
Commented by qaz last updated on 07/Jul/21
$$??? \\ $$
Answered by Kamel last updated on 07/Jul/21
$$ \\ $$$${S}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−{e}^{{x}} \right)^{{k}} =\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} {e}^{\left({n}+\mathrm{1}\right){x}} }{\mathrm{1}+{e}^{{x}} } \\ $$$${S}\left({x}\right)=\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} +\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right){x}+\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}{x}^{\mathrm{2}} +\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{6}}{x}^{\mathrm{3}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{48}}\right)+{o}\left({x}^{\mathrm{3}} \right) \\ $$$$\:\:\therefore\:\frac{{d}^{\mathrm{3}} {S}\left({x}\right)}{{dx}^{\mathrm{3}} }\mid_{{x}=\mathrm{0}} =\mathrm{6}\left(\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\mathrm{48}}−\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{8}}+\frac{\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{12}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}}{\mathrm{8}}−\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{3}}{\mathrm{4}}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)+\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\left({n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}{n}^{\mathrm{3}} +\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}}{n}^{\mathrm{2}} −\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\:\:\:\therefore\:\:\:\underset{\boldsymbol{{k}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\boldsymbol{\sum}}}\left(−\mathrm{1}\right)^{\boldsymbol{{k}}} \boldsymbol{{k}}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{8}}\left(\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \left(\mathrm{4}\boldsymbol{{n}}^{\mathrm{3}} +\mathrm{6}\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}\right) \\ $$$$ \\ $$