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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
Evaluate∑2n−1k=1(−1)k−1k3
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....
letS(x)=∑k=0N(−1)kxkwehaveS′(x)=∑k=1N(−1)kkxk−1⇒xS′(x)=∑k=1n(−1)kkxk⇒(xS′(x))′=∑k=1N(−1)kk2xk−1⇒S′(x)+xS″(x)=∑k=1N(−1)kk2xk−1⇒xS′(x)+x2S″(x)=∑k=1N(−1)kk2xk⇒(xS′(x)+x2S″(x))′=∑k=1N(−1)kk3xk−1⇒S′(x)+xS″(x)+2xS″(x)+x2S(3)(x)=∑k=1N(−1)kk3xk−1⇒xS′(x)+3x2S(2)(x)+x3S(3)(x)=∑k=1N(−1)kk3xklettakex=1⇒∑k=1N(−1)kxk=S′(1)+3S(2)(1)+S(3)(1)butS(x)=∑k=0N(−x)k=1−(−x)N+11+x=1−(−1)N+1xN+11+x(x≠−1)=1+(−1)NxN+11+x⇒S′(x)=(N+1)(−1)NxN(1+x)−(1+(−1)NxN+1)(1+x)2=(N+1)(−1)NxN+(N+1)(−1)NxN+1−1−(−1)NxN+1(1+x)2=(N+1)(−1)NxN+N(−1)NxN+1−1(1+x)2⇒S′(1)=(N+1)(−1)N+N(−1)N4=(2N+1)(−1)N4alsowemustcalculateS(2)(1)andS(3)(1)andtakeN=2n−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)
Sn=∑2n−1k=1(−1)k−1k3=1−23+33−43+…+(2n−1)3=(1+33+53+…+(2n−1)3)−(23+43+63+…+(2n−2)3)=[1+23+33+43+…+(2n−1)3−(23+43+63+…+(2n−2)3)]−23(13+23+33+…+(n−1)3)=∑2n−1k=1k3−23×2∑n−1k=1k3=[(2n−1)(2n−1+1)2]2−24[(n−1)(n−1+1)2]2note:∑nk=1k3=[n(n+1)2]2Sn=(2n−1)2(2n)222−24(n−1)2n222=n2((2n−1)2−22(n−1)2)=n2(4n2−4n+1−4n2+8n−4)Sn=n2(4n−3)

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