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Question Number 106314 by Study last updated on 04/Aug/20

Answered by bemath last updated on 04/Aug/20

GP ⇒S = (1/(1−(−1))) = (1/2)  1+1.(−1)+1(−1)^2 +1(−1)^3 +...

$$\mathrm{GP}\:\Rightarrow\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{1}−\left(−\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{1}.\left(−\mathrm{1}\right)+\mathrm{1}\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\left(−\mathrm{1}\right)^{\mathrm{3}} +... \\ $$

Commented by Study last updated on 04/Aug/20

∣q∣=1 for this form we have s_∞ =((a(q^n −1)/(q−1))

$$\mid{q}\mid=\mathrm{1}\:{for}\:{this}\:{form}\:{we}\:{have}\:{s}_{\infty} =\frac{{a}\left({q}^{{n}} −\mathrm{1}\right.}{{q}−\mathrm{1}} \\ $$

Answered by Dwaipayan Shikari last updated on 04/Aug/20

S=1+x+x^2 +x^3 +x^4 +x^5 +x^6 +x^7 +...  S=(1/(1−x))  S=1−1+1−1+1−1+1−1+1−1+1−1+...  =(1/(1+1))=(1/2)  It is  a Mean of the probable value

$${S}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +{x}^{\mathrm{6}} +{x}^{\mathrm{7}} +... \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${S}=\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+... \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${It}\:{is}\:\:{a}\:{Mean}\:{of}\:{the}\:{probable}\:{value} \\ $$

Answered by Her_Majesty last updated on 04/Aug/20

sum doesn′t exist

$${sum}\:{doesn}'{t}\:{exist} \\ $$

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