S-x-2x-2-nx-n-

Question Number 153847 by mathdanisur last updated on 11/Sep/21

$$\boldsymbol{\mathrm{S}}\:=\:\mathrm{x}\:+\:\mathrm{2x}^{\mathrm{2}} \:+\:…\:+\:\mathrm{nx}^{\boldsymbol{\mathrm{n}}} \\ $$$$ \\ $$

Answered by liberty last updated on 11/Sep/21

S=x+2x^2 +3x^3 +4x^4 +…+nx^n xS= x^2 +2x^3 +3x^4 +…+(n−1)x^n +nx^(n+1) (1−x)S=x+x^2 +x^3 +x^4 +…+x^n −nx^(n+1) (1−x)S=((x(1−x^n ))/(1−x))−nx^(n+1) S=((x(1−x^n ))/((1−x)^2 ))−((nx^(n+1) )/((1−x)))

$$\:{S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +\ldots+{nx}^{{n}} \\ $$$${xS}=\:\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} +\ldots+\left({n}−\mathrm{1}\right){x}^{{n}} +{nx}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{x}\right){S}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +\ldots+{x}^{{n}} −{nx}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{x}\right){S}=\frac{{x}\left(\mathrm{1}−{x}^{{n}} \right)}{\mathrm{1}−{x}}−{nx}^{{n}+\mathrm{1}} \\ $$$${S}=\frac{{x}\left(\mathrm{1}−{x}^{{n}} \right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{{nx}^{{n}+\mathrm{1}} }{\left(\mathrm{1}−{x}\right)} \\ $$

Commented by mathdanisur last updated on 11/Sep/21

$$\mathrm{thanks}\:\mathrm{ser}\:\mathrm{nice} \\ $$

Commented by peter frank last updated on 11/Sep/21

$$\mathrm{thanks} \\ $$

Answered by mr W last updated on 11/Sep/21

Method 1: x+x^2 +...+x^n =((x^n −1)/(x−1)) 1+2x+...+nx^(n−1) =((nx^(n−1) )/(x−1))−((x^n −1)/((x−1)^2 )) x+2x^2 +...+nx^n =((nx^n )/(x−1))−(((x^n −1)x)/((x−1)^2 )) Method 2: S=x+2x^2 +3x^3 +...+nx^n xS= x^2 +2x^3 +...+(n−1)x^n +nx^(n+1) (1−x)S=x+x^2 +x^3 +...x^n −nx^(n+1) (1−x)S=(((x^n −1)x)/(x−1))−nx^(n+1) ⇒S=((nx^(n+1) )/(x−1))−(((x^n −1)x)/((x−1)^2 ))

$${Method}\:\mathrm{1}: \\ $$$${x}+{x}^{\mathrm{2}} +…+{x}^{{n}} =\frac{{x}^{{n}} −\mathrm{1}}{{x}−\mathrm{1}} \\ $$$$\mathrm{1}+\mathrm{2}{x}+…+{nx}^{{n}−\mathrm{1}} =\frac{{nx}^{{n}−\mathrm{1}} }{{x}−\mathrm{1}}−\frac{{x}^{{n}} −\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}+\mathrm{2}{x}^{\mathrm{2}} +…+{nx}^{{n}} =\frac{{nx}^{{n}} }{{x}−\mathrm{1}}−\frac{\left({x}^{{n}} −\mathrm{1}\right){x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${Method}\:\mathrm{2}: \\ $$$${S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +…+{nx}^{{n}} \\ $$$${xS}=\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +…+\left({n}−\mathrm{1}\right){x}^{{n}} +{nx}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{x}\right){S}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…{x}^{{n}} −{nx}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{x}\right){S}=\frac{\left({x}^{{n}} −\mathrm{1}\right){x}}{{x}−\mathrm{1}}−{nx}^{{n}+\mathrm{1}} \\ $$$$\Rightarrow{S}=\frac{{nx}^{{n}+\mathrm{1}} }{{x}−\mathrm{1}}−\frac{\left({x}^{{n}} −\mathrm{1}\right){x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by mathdanisur last updated on 11/Sep/21

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{thanks}\:\mathrm{ser} \\ $$

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