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