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Question Number 2258 by Filup last updated on 12/Nov/15
Solve the following sum:  Σ_(i=1) ^∞ (−1)^(i+1) ix^i =x−2x^2 +3x^3 −4x^4 +...
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{sum}: \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}+\mathrm{1}} {ix}^{{i}} ={x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{4}} +… \\ $$
Commented by Filup last updated on 12/Nov/15
For: ∣x∣<1
$$\mathrm{For}:\:\mid{x}\mid<\mathrm{1} \\ $$
Answered by prakash jain last updated on 12/Nov/15
S  =x−2x^2 +3x^3 −4x^4 +...                (1)  xS=  +x^2 −2x^3 +3x^4 −....                 (2)  add (1) and (2)  (1+x)S=x−x^2 +x^3 −x^4 +...=(x/(1+x))  S=(x/((1+x)^2 ))
$$\mathrm{S}\:\:={x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{4}} +…\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$${xS}=\:\:+{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} −….\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\mathrm{add}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}+{x}\right){S}={x}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{x}^{\mathrm{4}} +…=\frac{{x}}{\mathrm{1}+{x}} \\ $$$${S}=\frac{{x}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} } \\ $$
Commented by Filup last updated on 12/Nov/15
Wow! I couldnt work this out!  I′ll have to remember that!
$$\mathrm{Wow}!\:\mathrm{I}\:\mathrm{couldnt}\:\mathrm{work}\:\mathrm{this}\:\mathrm{out}! \\ $$$$\mathrm{I}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{remember}\:\mathrm{that}! \\ $$
Commented by 123456 last updated on 14/Nov/15
x=1  1−2+3−4+∙∙∙=(1/((1+1)^2 ))=(1/4)  XD
$${x}=\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\centerdot\centerdot\centerdot=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\:\:\mathrm{XD} \\ $$
Commented by prakash jain last updated on 14/Nov/15
1−1+1−1+1−1+....=(1/2)?
$$\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{1}−\mathrm{1}+….=\frac{\mathrm{1}}{\mathrm{2}}? \\ $$

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