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Question Number 41454 by Fawomath last updated on 07/Aug/18
Evaluate Σ_(k=1) ^(2n−1) (−1)^(k−1) k^3
$$\mathrm{Evaluate}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {k}^{\mathrm{3}} \\ $$
Commented by maxmathsup by imad last updated on 07/Aug/18
let S(x)=Σ_(k=0) ^N (−1)^k x^k we have S^′ (x)=Σ_(k=1) ^N (−1)^k k x^(k−1) ⇒ xS^′ (x)=Σ_(k=1) ^n (−1)^k k x^k ⇒ (xS^′ (x))^′ =Σ_(k=1) ^N (−1)^k k^2 x^(k−1) ⇒ S^′ (x) +x S^(′′) (x) =Σ_(k=1) ^N (−1)^k k^2 x^(k−1) ⇒ xS^′ (x) +x^2 S^(′′) (x) =Σ_(k=1) ^N (−1)^k k^2 x^k ⇒ (xS^′ (x) +x^2 S^(′′) (x))^′ = Σ_(k=1) ^N (−1)^k k^3 x^(k−1) ⇒ S^′ (x) +xS^(′′) (x) +2x S^(′′) (x) +x^2 S^((3)) (x) = Σ_(k=1) ^N (−1)^k k^3 x^(k−1) ⇒ xS^′ (x) +3x^2 S^((2)) (x) +x^3 S^((3)) (x) =Σ_(k=1) ^N (−1)^k k^3 x^(k ) let take x=1 ⇒ Σ_(k=1) ^N (−1)^k x^k = S^′ (1) +3 S^((2)) (1) +S^((3)) (1) but S(x)=Σ_(k=0) ^N (−x)^k =((1−(−x)^(N+1) )/(1+x)) = ((1 −(−1)^(N+1) x^(N+1) )/(1+x)) (x≠−1) = ((1+(−1)^N x^(N+1) )/(1+x)) ⇒ S^′ (x) =(((N+1)(−1)^N x^N (1+x) −(1+(−1)^N x^(N+1) ))/((1+x)^2 )) =(((N+1)(−1)^N x^N +(N+1)(−1)^N x^(N+1) −1−(−1)^N x^(N+1) )/((1+x)^2 )) =(((N+1)(−1)^N x^N +N(−1)^N x^(N+1) −1)/((1+x^ )^2 )) ⇒ S^′ (1)=(((N+1)(−1)^N +N(−1)^N )/4) =(((2N+1)(−1)^N )/4) also we must calculate S^((2)) (1) and S^((3)) (1) and take N =2n−1....
$${let}\:{S}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}} \:\:\:{we}\:{have}\:{S}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{N}} \left(−\mathrm{1}\right)^{{k}} {k}\:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${xS}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} {k}\:{x}^{{k}} \:\Rightarrow\:\left({xS}^{'} \left({x}\right)\right)^{'} \:=\sum_{{k}=\mathrm{1}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} \:{k}^{\mathrm{2}} \:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${S}^{'} \left({x}\right)\:+{x}\:{S}^{''} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{N}} \left(−\mathrm{1}\right)^{{k}} {k}^{\mathrm{2}} \:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${xS}^{'} \left({x}\right)\:+{x}^{\mathrm{2}} \:{S}^{''} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} {k}^{\mathrm{2}} {x}^{{k}} \:\Rightarrow \\ $$$$\left({xS}^{'} \left({x}\right)\:+{x}^{\mathrm{2}} {S}^{''} \left({x}\right)\right)^{'} =\:\sum_{{k}=\mathrm{1}} ^{{N}} \left(−\mathrm{1}\right)^{{k}} {k}^{\mathrm{3}} \:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${S}^{'} \left({x}\right)\:+{xS}^{''} \left({x}\right)\:+\mathrm{2}{x}\:{S}^{''} \left({x}\right)\:+{x}^{\mathrm{2}} \:{S}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\:\sum_{{k}=\mathrm{1}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} \:{k}^{\mathrm{3}} \:{x}^{{k}−\mathrm{1}} \:\Rightarrow \\ $$$${xS}^{'} \left({x}\right)\:\:+\mathrm{3}{x}^{\mathrm{2}} \:{S}^{\left(\mathrm{2}\right)} \left({x}\right)\:+{x}^{\mathrm{3}} \:{S}^{\left(\mathrm{3}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{1}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} \:{k}^{\mathrm{3}} \:{x}^{{k}\:\:} \:\:\:{let}\:{take}\:{x}=\mathrm{1}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{N}} \:\left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}} \:=\:{S}^{'} \left(\mathrm{1}\right)\:+\mathrm{3}\:{S}^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\:+{S}^{\left(\mathrm{3}\right)} \left(\mathrm{1}\right)\:\:{but} \\ $$$${S}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{N}} \left(−{x}\right)^{{k}} \:\:=\frac{\mathrm{1}−\left(−{x}\right)^{{N}+\mathrm{1}} }{\mathrm{1}+{x}}\:\:=\:\frac{\mathrm{1}\:−\left(−\mathrm{1}\right)^{{N}+\mathrm{1}} \:{x}^{{N}+\mathrm{1}} }{\mathrm{1}+{x}}\:\:\:\:\left({x}\neq−\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}+\mathrm{1}} }{\mathrm{1}+{x}}\:\Rightarrow\:{S}^{'} \left({x}\right)\:=\frac{\left({N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} {x}^{{N}} \left(\mathrm{1}+{x}\right)\:−\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{N}} {x}^{{N}+\mathrm{1}} \right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}} \:+\left({N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}+\mathrm{1}} \:−\mathrm{1}−\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}+\mathrm{1}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}} \:+{N}\left(−\mathrm{1}\right)^{{N}} \:{x}^{{N}+\mathrm{1}} \:−\mathrm{1}}{\left(\mathrm{1}+{x}^{} \right)^{\mathrm{2}} }\:\:\Rightarrow\:{S}^{'} \left(\mathrm{1}\right)=\frac{\left({N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} \:+{N}\left(−\mathrm{1}\right)^{{N}} }{\mathrm{4}} \\ $$$$=\frac{\left(\mathrm{2}{N}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{N}} }{\mathrm{4}}\:\:\:{also}\:{we}\:{must}\:{calculate}\:{S}^{\left(\mathrm{2}\right)} \left(\mathrm{1}\right)\:{and}\:{S}^{\left(\mathrm{3}\right)} \left(\mathrm{1}\right)\:{and} \\ $$$${take}\:{N}\:=\mathrm{2}{n}−\mathrm{1}…. \\ $$
Answered by sma3l2996 last updated on 07/Aug/18
S_n =Σ_(k=1) ^(2n−1) (−1)^(k−1) k^3 =1−2^3 +3^3 −4^3 +...+(2n−1)^3 =(1+3^3 +5^3 +...+(2n−1)^3 )−(2^3 +4^3 +6^3 +...+(2n−2)^3 ) =[1+2^3 +3^3 +4^3 +...+(2n−1)^3 −(2^3 +4^3 +6^3 +...+(2n−2)^3 )]−2^3 (1^3 +2^3 +3^3 +...+(n−1)^3 ) =Σ_(k=1) ^(2n−1) k^3 −2^3 ×2Σ_(k=1) ^(n−1) k^3 =[(((2n−1)(2n−1+1))/2)]^2 −2^4 [(((n−1)(n−1+1))/2)]^2 note: Σ_(k=1) ^n k^3 =[((n(n+1))/2)]^2 S_n =(((2n−1)^2 (2n)^2 )/2^2 )−2^4 (((n−1)^2 n^2 )/2^2 ) =n^2 ((2n−1)^2 −2^2 (n−1)^2 ) =n^2 (4n^2 −4n+1−4n^2 +8n−4) S_n =n^2 (4n−3)
$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {k}^{\mathrm{3}} \\ $$$$=\mathrm{1}−\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} −\mathrm{4}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{1}+\mathrm{3}^{\mathrm{3}} +\mathrm{5}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} \right)−\left(\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{2}\right)^{\mathrm{3}} \right) \\ $$$$=\left[\mathrm{1}+\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} −\left(\mathrm{2}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} +\mathrm{6}^{\mathrm{3}} +…+\left(\mathrm{2}{n}−\mathrm{2}\right)^{\mathrm{3}} \right)\right]−\mathrm{2}^{\mathrm{3}} \left(\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+\left({n}−\mathrm{1}\right)^{\mathrm{3}} \right) \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}{k}^{\mathrm{3}} −\mathrm{2}^{\mathrm{3}} ×\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{k}^{\mathrm{3}} \\ $$$$=\left[\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} −\mathrm{2}^{\mathrm{4}} \left[\frac{\left({n}−\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$${note}:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{3}} =\left[\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$${S}_{{n}} =\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} }−\mathrm{2}^{\mathrm{4}} \frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} {n}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} } \\ $$$$={n}^{\mathrm{2}} \left(\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$={n}^{\mathrm{2}} \left(\mathrm{4}{n}^{\mathrm{2}} −\mathrm{4}{n}+\mathrm{1}−\mathrm{4}{n}^{\mathrm{2}} +\mathrm{8}{n}−\mathrm{4}\right) \\ $$$${S}_{{n}} ={n}^{\mathrm{2}} \left(\mathrm{4}{n}−\mathrm{3}\right) \\ $$

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