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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) )

$$\mathrm{S}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \\ $$$${S}=\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{x}^{\mathrm{4}} −{x}^{\mathrm{5}} +..−{x}^{\mathrm{2}{n}−\mathrm{1}} +{x}^{\mathrm{2}{n}} \\ $$$${xS}=\:\:\:{x}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{x}^{\mathrm{4}} +−...+{x}^{\mathrm{2}{n}−\mathrm{1}} −{x}^{\mathrm{2}{n}} +{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}+{x}\right){S}=\left(\mathrm{1}+{x}^{\mathrm{2}{n}+\mathrm{1}} \right) \\ $$$${or} \\ $$$$\left(\mathrm{1}+{x}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} =\left(\mathrm{1}+{x}^{\mathrm{2}{n}+\mathrm{1}} \right) \\ $$

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.

$${let}\:{put}\:{S}_{{n}} =\:\left({x}+\mathrm{1}\right)\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \:\:{with}\:{n}\:{from}\:{N} \\ $$$${if}\:{x}=−\mathrm{1}\:\:{S}_{{n}} =\:\mathrm{0}=\:\left(−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{1} \\ $$$${if}\:{x}\neq−\mathrm{1}\:\:\:{S}_{{n}} =\left({x}+\mathrm{1}\right)\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \left(−{x}\right)^{{k}} \\ $$$$=\left({x}+\mathrm{1}\right)\:\frac{\mathrm{1}−\left(−{x}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−\left(−{x}\right)}=\:\:\mathrm{1}+{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${in}\:{the}\:{case}\:{nfrom}\:{Z}^{−} \:{we}\:{use}\:{the}\:{ch}.\:{n}=−{p}\:{we}\:{find}\:{the}\:{same}\:{resuls}. \\ $$$$ \\ $$

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