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Question Number 27376 by macanudo last updated on 05/Jan/18

Commented by prakash jain last updated on 05/Jan/18

S=Σ_(k=0) ^(2n) (−1)^k x^k S=1−x+x^2 −x^3 +x^4 −x^5 +..−x^(2n−1) +x^(2n) xS= x−x^2 +x^3 −x^4 +−...+x^(2n−1) −x^(2n) +x^(2n+1) (1+x)S=(1+x^(2n+1) ) or (1+x)Σ_(k=0) ^(2n) (−1)^k x^k =(1+x^(2n+1) )

S=∑2nk=0(−1)kxkS=1−x+x2−x3+x4−x5+..−x2n−1+x2nxS=x−x2+x3−x4+−...+x2n−1−x2n+x2n+1(1+x)S=(1+x2n+1)or(1+x)∑2nk=0(−1)kxk=(1+x2n+1)

Commented by abdo imad last updated on 06/Jan/18

let put S_n = (x+1) Σ_(k=0) ^(2n) (−1)^k x^k with n from N if x=−1 S_n = 0= (−1)^(2n+1) +1 if x≠−1 S_n =(x+1) Σ_(k=0) ^(2n) (−x)^k =(x+1) ((1−(−x)^(2n+1) )/(1−(−x)))= 1+x^(2n+1) in the case nfrom Z^− we use the ch. n=−p we find the same resuls.

letputSn=(x+1)∑k=02n(−1)kxkwithnfromNifx=−1Sn=0=(−1)2n+1+1ifx≠−1Sn=(x+1)∑k=02n(−x)k=(x+1)1−(−x)2n+11−(−x)=1+x2n+1inthecasenfromZ−weusethech.n=−pwefindthesameresuls.

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