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Question-187546

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Question Number 187546 by yaslm last updated on 18/Feb/23
Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23
S=((1×2)/3)+((2×3)/3^2 )+((3×4)/3^3 )+((4×5)/3^4 )+∙∙∙ (i) 3S=1×2+((2×3)/3)+((3×4)/3^2 )+((4×5)/3^3 )+∙∙∙ (ii) (ii)−(i) 2S=2+((2×2)/3)+((2×3)/3^2 )+((2×4)/3^3 )+((2×5)/3^4 )+∙∙∙ (iii) 6S=2×3+2×2+((2×3)/3)+((2×4)/3^2 )+((2×5)/3^3 )+∙∙∙ (iv) (iv)−(iii) 4S=8+(2/3)+(2/3^2 )+(2/3^3 )+(2/3^4 )+∙∙∙ (v) 12S=24+2+(2/3)+(2/3^2 )+(2/3^3 )+(2/3^4 )+∙∙∙ (vi) 12S−4S=18 ⇒8S=18 ⇒S=(9/4)
$${S}=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}×\mathrm{4}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}×\mathrm{5}}{\mathrm{3}^{\mathrm{4}} }+\centerdot\centerdot\centerdot\:\:\left({i}\right) \\ $$$$\mathrm{3}{S}=\mathrm{1}×\mathrm{2}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}}+\frac{\mathrm{3}×\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{4}×\mathrm{5}}{\mathrm{3}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\:\:\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right) \\ $$$$\mathrm{2}{S}=\mathrm{2}+\frac{\mathrm{2}×\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}×\mathrm{4}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{2}×\mathrm{5}}{\mathrm{3}^{\mathrm{4}} }+\centerdot\centerdot\centerdot\:\left({iii}\right) \\ $$$$\mathrm{6}{S}=\mathrm{2}×\mathrm{3}+\mathrm{2}×\mathrm{2}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}}+\frac{\mathrm{2}×\mathrm{4}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}×\mathrm{5}}{\mathrm{3}^{\mathrm{3}} }+\centerdot\centerdot\centerdot\:\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right) \\ $$$$\mathrm{4}{S}=\mathrm{8}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{4}} }+\centerdot\centerdot\centerdot\:\:\left({v}\right) \\ $$$$\mathrm{12}{S}=\mathrm{24}+\mathrm{2}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{4}} }+\centerdot\centerdot\centerdot\:\:\left({vi}\right) \\ $$$$\mathrm{12}{S}−\mathrm{4}{S}=\mathrm{18}\:\Rightarrow\mathrm{8}{S}=\mathrm{18}\:\Rightarrow{S}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Answered by JDamian last updated on 18/Feb/23
S = Σ_(k=1) ^(∞) ((k(k+1))/3^k ) f ≡ f(x) = 1+x+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 g(x)≡D{x^2 Df} = 1∙2x+2∙3x^2 +3∙4x^3 +∙∙∙= = ((2x)/((1−x)^3 )) ∀∣x∣<1 S=g(x)∣_(x=(1/3)) =g((1/3))= =((2×(1/3))/((1−(1/3))^3 ))=(2/(3×((2/3))^3 ))=(2/(3×((2×2^2 )/(3×3^2 )) ))=(9/4)■
$${S}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\Sigma}}\frac{{k}\left({k}+\mathrm{1}\right)}{\mathrm{3}^{{k}} } \\ $$$${f}\:\equiv\:{f}\left({x}\right)\:=\:\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +\:\centerdot\centerdot\centerdot\:=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${D}\:\equiv\:\frac{{d}}{{dx}} \\ $$$${Df}\:=\:\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +\:\centerdot\centerdot\centerdot\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${x}^{\mathrm{2}} {Df}\:=\:\mathrm{1}\centerdot{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\:\centerdot\centerdot\centerdot\:=\:\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$${g}\left({x}\right)\equiv{D}\left\{{x}^{\mathrm{2}} {Df}\right\}\:=\:\mathrm{1}\centerdot\mathrm{2}{x}+\mathrm{2}\centerdot\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}\centerdot\mathrm{4}{x}^{\mathrm{3}} +\centerdot\centerdot\centerdot= \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\forall\mid{x}\mid<\mathrm{1} \\ $$$$ \\ $$$${S}={g}\left({x}\right)\mid_{{x}=\frac{\mathrm{1}}{\mathrm{3}}} ={g}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)= \\ $$$$=\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\mathrm{2}}{\mathrm{3}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} }=\frac{\cancel{\mathrm{2}}}{\cancel{\mathrm{3}}×\frac{\cancel{\mathrm{2}}×\mathrm{2}^{\mathrm{2}} }{\cancel{\mathrm{3}}×\mathrm{3}^{\mathrm{2}} }\:}=\frac{\mathrm{9}}{\mathrm{4}}\blacksquare \\ $$

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