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Question Number 61724 by aliesam last updated on 07/Jun/19

Σ_(n≥0) n^2 x^n

$$\underset{{n}\geqslant\mathrm{0}} {\sum}{n}^{\mathrm{2}} {x}^{{n}} \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 07/Jun/19

let p(x) =Σ_(n=0) ^∞ x^n (∣x∣<1) ⇒p(x) =(1/(1−x)) ⇒p^′ (x) =(1/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒Σ_(n=1) ^∞ nx^n =(x/((x−1)^2 )) ⇒ Σ_(n=1) ^∞ n^2 x^(n−1) ={ (x/((x−1)^2 ))}^((1)) =(((x−1)^2 −x(2(x−1)))/((x−1)^4 )) =((x^2 −2x+1−2x^2 +2x)/((x−1)^4 )) =((1−x^2 )/((1−x)^4 )) =(((1−x)(1+x))/((1−x)^4 )) =((x+1)/((1−x)^3 )) ⇒Σ_(n=1) ^∞ n^2 x^n =((x^2 +x)/((1−x)^3 )) ⇒ Σ_(n≥0) n^2 x^n = ((x^2 +x)/((1−x)^3 )) .

$${let}\:{p}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:\:\left(\mid{x}\mid<\mathrm{1}\right)\:\Rightarrow{p}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{p}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:=\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \:=\left\{\:\frac{{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \:=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −{x}\left(\mathrm{2}\left({x}−\mathrm{1}\right)\right)}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} }\:=\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} \:{x}^{{n}} \:=\frac{{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}\geqslant\mathrm{0}} \:{n}^{\mathrm{2}} {x}^{{n}} \:\:=\:\frac{{x}^{\mathrm{2}} +{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:. \\ $$

Commented by maxmathsup by imad last updated on 07/Jun/19

sir aliissam where are you from....

$${sir}\:\:{aliissam}\:{where}\:{are}\:{you}\:{from}.... \\ $$

Commented by aliesam last updated on 07/Jun/19

iraq sir

$${iraq}\:{sir}\: \\ $$

Commented by aliesam last updated on 07/Jun/19

but why

$${but}\:{why}\: \\ $$

Commented by maxmathsup by imad last updated on 07/Jun/19

for nothing its only a information sir ...hou are always welcome...

$${for}\:{nothing}\:{its}\:{only}\:{a}\:{information}\:{sir}\:...{hou}\:{are}\:{always}\:{welcome}... \\ $$

Commented by aliesam last updated on 07/Jun/19

ok sir and thank you about the graet sol

$${ok}\:{sir}\:{and}\:{thank}\:{you}\:{about}\:{the}\:{graet}\:{sol} \\ $$

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