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Question Number 106488 by Algoritm last updated on 05/Aug/20
Commented by Dwaipayan Shikari last updated on 05/Aug/20
Q106379 ((1.2^2 )/(10))+((2.3^2 )/(10))+.....=Σ_(n=1) ^∞ ((n(n+1)^2 )/(10^n ))
$$\:{Q}\mathrm{106379} \\ $$$$\frac{\mathrm{1}.\mathrm{2}^{\mathrm{2}} }{\mathrm{10}}+\frac{\mathrm{2}.\mathrm{3}^{\mathrm{2}} }{\mathrm{10}}+…..=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{n}} } \\ $$
Answered by abdomsup last updated on 05/Aug/20
S=Σ_(n=1) ^∞ ((n(n+1)^2 )/(10^n )) =Σ_(n=1) ^∞ ((n(n^2 +2n+1))/(10^n )) =Σ_(n=0) ^∞ (n^3 /(10^n )) +2Σ_(n=0) ^∞ (n^2 /(10^n )) +Σ_(n=0) ^∞ (1/(10^n )) we have for ∣x∣<1 Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒Σ_(n=0) ^∞ (1/(10^n )) =(1/(1−(1/(10)))) =((10)/9) also Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ n^2 x^(n−1 ) =(((x−1)^2 −2(x−1)x)/((x−1)^4 )) =((x−1−2x)/((x−1)^3 )) =((−x−1)/((x−1)^3 )) =((1+x)/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ n^2 x^n =((x+x^2 )/((1−x)^3 )) ⇒ Σ_(n=1) ^∞ (n^2 /(10^n )) =(((1/(10))+((1/(10)))^2 )/((1−(1/(10)))^3 )) =... Σ_(n=1) ^∞ n^3 x^(n−1) =(((2x+1)(1−x)^3 +3(1−x)^2 (x^2 +x))/((1−x)^6 )) =(((2x+1)(1−x)+3x^2 +3x)/((1−x)^4 )) =((2x+1−2x^2 −x+3x^2 +3x)/((1−x)^4 )) =((x^2 +4x+1)/((1−x)^4 )) ⇒ Σ_(n=1) ^∞ n^3 x^n =((x^3 +4x^2 +x)/((1−x)^4 )) ⇒ Σ_(n=1) ^(∞ ) (n^3 /(10^n )) =(((1/(10^3 ))+(4/(10^2 ))+(1/(10)))/((1−(1/(10)))^4 ))=...
$${S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{n}} } \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{10}^{{n}} } \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}^{\mathrm{3}} }{\mathrm{10}^{{n}} }\:+\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}^{\mathrm{2}} }{\mathrm{10}^{{n}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{10}^{{n}} } \\ $$$${we}\:{have}\:{for}\:\mid{x}\mid<\mathrm{1}\: \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{10}^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}\:=\frac{\mathrm{10}}{\mathrm{9}} \\ $$$${also}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}\:} =\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left({x}−\mathrm{1}\right){x}}{\left({x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}−\mathrm{1}−\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{−{x}−\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}} \:=\frac{{x}+{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{2}} }{\mathrm{10}^{{n}} }\:=\frac{\frac{\mathrm{1}}{\mathrm{10}}+\left(\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{3}} }\:=… \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} {n}^{\mathrm{3}} {x}^{{n}−\mathrm{1}} \:=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} } \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{\mathrm{2}{x}+\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} \:{x}^{{n}} \:=\frac{{x}^{\mathrm{3}} \:+\mathrm{4}{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty\:} \:\frac{{n}^{\mathrm{3}} }{\mathrm{10}^{{n}} }\:=\frac{\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{10}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{4}} }=… \\ $$
Answered by Dwaipayan Shikari last updated on 05/Aug/20
Σ^∞ ((n(n+1)^2 )/(10^n )) Σ^∞ (n^3 /(10^n ))+Σ^∞ ((2n^2 )/(10^n ))+Σ^∞ (n/(10^n )) Σ^∞ (n/(10^n ))=(1/(10))+(2/(10^2 ))+...=S (S/(10))=(1/(10^2 ))+(2/(10^3 ))+.... ((9S)/(10))=(1/(10))+(1/(10^2 ))+(1/(10^3 ))+... ((9S)/(10))=((1/(10))/(1−(1/(10))))=(1/9) S=((10)/(81)) 2Σ^∞ (n^2 /(10^n ))=(1/(10))+(4/(10^2 ))+(9/(10^2 ))+....=S′ ((S′)/(10))=(1/(10^2 ))+(4/(10^3 ))+(9/(10^4 ))+... ((9S′)/(10))=(1/(10))+(3/(10^2 ))+(5/(10^3 ))+... ((9S^′ )/(100))= (1/(10^2 ))+(3/(10^3 ))+... ......subtracting ((9S′)/(10))(1−(1/(10)))=(1/(10))+2((1/(10^2 ))+.....) ((81S′)/(100))=(2/9)−(1/(10)) S′=((10)/9).((11)/(81))=((110)/(729)) Σ^∞ ((n^3 /(10^n )))=(1/(10))+(8/(10^2 ))+((27)/(10^3 ))+...=S_n S_n +2S′+S=Σ^∞ (n^3 /(10^n ))+((220)/(729))+((10)/(81))=((1400)/(2187))
$$\overset{\infty} {\sum}\frac{{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{n}} } \\ $$$$\overset{\infty} {\sum}\frac{{n}^{\mathrm{3}} }{\mathrm{10}^{{n}} }+\overset{\infty} {\sum}\frac{\mathrm{2}{n}^{\mathrm{2}} }{\mathrm{10}^{{n}} }+\overset{\infty} {\sum}\frac{{n}}{\mathrm{10}^{{n}} } \\ $$$$\overset{\infty} {\sum}\frac{{n}}{\mathrm{10}^{{n}} }=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{2}}{\mathrm{10}^{\mathrm{2}} }+…={S} \\ $$$$\frac{{S}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{10}^{\mathrm{3}} }+…. \\ $$$$\frac{\mathrm{9}{S}}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{3}} }+… \\ $$$$\frac{\mathrm{9}{S}}{\mathrm{10}}=\frac{\frac{\mathrm{1}}{\mathrm{10}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}}=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${S}=\frac{\mathrm{10}}{\mathrm{81}} \\ $$$$\mathrm{2}\overset{\infty} {\sum}\frac{{n}^{\mathrm{2}} }{\mathrm{10}^{{n}} }=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{4}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{9}}{\mathrm{10}^{\mathrm{2}} }+….={S}' \\ $$$$\frac{{S}'}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{10}^{\mathrm{3}} }+\frac{\mathrm{9}}{\mathrm{10}^{\mathrm{4}} }+… \\ $$$$\frac{\mathrm{9}{S}'}{\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{3}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{5}}{\mathrm{10}^{\mathrm{3}} }+… \\ $$$$\frac{\mathrm{9}{S}^{'} }{\mathrm{100}}=\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{10}^{\mathrm{3}} }+… \\ $$$$……{subtracting} \\ $$$$\frac{\mathrm{9}{S}'}{\mathrm{10}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{10}}\right)=\frac{\mathrm{1}}{\mathrm{10}}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{2}} }+…..\right) \\ $$$$\frac{\mathrm{81}{S}'}{\mathrm{100}}=\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${S}'=\frac{\mathrm{10}}{\mathrm{9}}.\frac{\mathrm{11}}{\mathrm{81}}=\frac{\mathrm{110}}{\mathrm{729}} \\ $$$$\overset{\infty} {\sum}\left(\frac{{n}^{\mathrm{3}} }{\mathrm{10}^{{n}} }\right)=\frac{\mathrm{1}}{\mathrm{10}}+\frac{\mathrm{8}}{\mathrm{10}^{\mathrm{2}} }+\frac{\mathrm{27}}{\mathrm{10}^{\mathrm{3}} }+…={S}_{{n}} \\ $$$${S}_{{n}} +\mathrm{2}{S}'+{S}=\overset{\infty} {\sum}\frac{{n}^{\mathrm{3}} }{\mathrm{10}^{{n}} }+\frac{\mathrm{220}}{\mathrm{729}}+\frac{\mathrm{10}}{\mathrm{81}}=\frac{\mathrm{1400}}{\mathrm{2187}} \\ $$
Answered by JDamian last updated on 06/Aug/20
S = Σ_(k=1) ^(∞) ((k(k+1)^2 )/(10^k )) f ≡ f(x) = 1+x^2 +x^3 + ∙∙∙ = (1/(1−x)) ∀∣x∣<1 D ≡ (d/dx) Df = 1+2x+3x^2 + ∙∙∙ = (1/((1−x)^2 )) ∀∣x∣<1 x^2 Df = 1∙x^2 +2x^3 + ∙∙∙ = (x^2 /((1−x)^2 )) ∀∣x∣<1 D(x^2 Df) = 1∙2x+2∙3x^2 +3∙4x^3 + ∙∙∙ = = 2(x/((1−x)^3 )) ∀∣x∣<1 xD(x^2 Df) = 1∙2x^2 +2∙3x^3 +3∙4x^4 + ∙∙∙ = = 2(x^2 /((1−x)^3 )) ∀∣x∣<1 g≡D(xD(x^2 Df)) = 1∙2^2 x+2∙3^2 x^2 + +3∙4^2 x^3 +4∙5^2 x^4 + ∙∙∙ = = 2(((2+x)x)/((1−x)^4 )) ∀∣x∣<1 g((1/(10)))=2(((2+(1/(10)))(1/(10)))/(((9/(10)))^4 ))=2(((21)/(100))/(((9/(10)))^4 ))=((4200)/9^4 )=((4200)/(6561))
$${S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\frac{{k}\left({k}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{10}^{{k}} } \\ $$$${f}\:\equiv\:{f}\left({x}\right)\:=\:\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +\:\centerdot\centerdot\centerdot\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${D}\:\equiv\:\frac{{d}}{{dx}} \\ $$$${Df}\:=\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${x}^{\mathrm{2}} {Df}\:=\:\mathrm{1}\centerdot{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\:\centerdot\centerdot\centerdot\:=\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${D}\left({x}^{\mathrm{2}} {Df}\right)\:=\:\mathrm{1}\centerdot\mathrm{2}{x}+\mathrm{2}\centerdot\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}\centerdot\mathrm{4}{x}^{\mathrm{3}} +\:\centerdot\centerdot\centerdot\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\:\:\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${xD}\left({x}^{\mathrm{2}} {Df}\right)\:=\:\mathrm{1}\centerdot\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\centerdot\mathrm{3}{x}^{\mathrm{3}} +\mathrm{3}\centerdot\mathrm{4}{x}^{\mathrm{4}} +\:\centerdot\centerdot\centerdot\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\:\:\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${g}\equiv{D}\left({xD}\left({x}^{\mathrm{2}} {Df}\right)\right)\:=\:\mathrm{1}\centerdot\mathrm{2}^{\mathrm{2}} {x}+\mathrm{2}\centerdot\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\centerdot\mathrm{4}^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{4}\centerdot\mathrm{5}^{\mathrm{2}} {x}^{\mathrm{4}} +\:\centerdot\centerdot\centerdot\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\frac{\left(\mathrm{2}+{x}\right){x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\:\:\:\:\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$$ \\ $$$${g}\left(\frac{\mathrm{1}}{\mathrm{10}}\right)=\mathrm{2}\frac{\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{10}}\right)\frac{\mathrm{1}}{\mathrm{10}}}{\left(\frac{\mathrm{9}}{\mathrm{10}}\right)^{\mathrm{4}} }=\mathrm{2}\frac{\frac{\mathrm{21}}{\mathrm{100}}}{\left(\frac{\mathrm{9}}{\mathrm{10}}\right)^{\mathrm{4}} }=\frac{\mathrm{4200}}{\mathrm{9}^{\mathrm{4}} }=\frac{\mathrm{4200}}{\mathrm{6561}} \\ $$

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