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Question Number 46852 by maxmathsup by imad last updated on 01/Nov/18
find the value of Σ_(n=1) ^∞ (n^3 /3^n )
$${find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} } \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
we have proved that for ∣x∣<1 Σ_(n=1) ^∞ nx^n =(x/((1−x)^2 )) ⇒ Σ_(n=1) ^∞ n^2 x^(n−1) =(((1−x)^2 +2x(1−x))/((1−x)^4 )) = ((1−x +2x)/((1−x)^3 ))=((x+1)/((1−x)^3 ))⇒ n^2 x^n =((x^2 +x)/((1−x)^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)+3(x^2 +x))/((1−x)^4 )) =((2x−2x^2 +1−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 )) x=(1/3) ⇒Σ_(n=1) ^∞ n^3 ((1/3))^n =(((1/(27))+(4/9)+(1/3))/(((2/3))^4 )) =(1/(27)) ((1 +12+9)/(2^4 /3^4 )) =(1/(27)) (3^4 /2^4 ) .22 =(3/(8.2)) .((11.2)/1) =((33)/8) ⇒ ★Σ_(n=1) ^∞ (n^3 /3^n ) =((33)/8) ★
$${we}\:{have}\:{proved}\:{that}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}} \:=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \:=\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\:\frac{\mathrm{1}−{x}\:+\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\frac{{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\Rightarrow \\ $$$$\:{n}^{\mathrm{2}} {x}^{{n}} \:\:=\frac{{x}^{\mathrm{2}} +{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} {x}^{{n}−\mathrm{1}} \:=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{6}} } \\ $$$$=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)+\mathrm{3}\left({x}^{\mathrm{2}} +{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}−{x}\:\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{3}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:=\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}\:+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} {x}^{{n}} \:=\frac{{x}^{\mathrm{3}} \:+\mathrm{4}{x}^{\mathrm{2}} \:+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \:=\frac{\frac{\mathrm{1}}{\mathrm{27}}+\frac{\mathrm{4}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{4}} }\:=\frac{\mathrm{1}}{\mathrm{27}}\:\frac{\mathrm{1}\:+\mathrm{12}+\mathrm{9}}{\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{3}^{\mathrm{4}} }}\:=\frac{\mathrm{1}}{\mathrm{27}}\:\frac{\mathrm{3}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} }\:.\mathrm{22} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}.\mathrm{2}}\:.\frac{\mathrm{11}.\mathrm{2}}{\mathrm{1}}\:\:=\frac{\mathrm{33}}{\mathrm{8}}\:\Rightarrow\:\bigstar\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }\:\:=\frac{\mathrm{33}}{\mathrm{8}}\:\bigstar \\ $$
Answered by Smail last updated on 02/Nov/18
(1/(1−x))=Σ_(n=0) ^∞ x^n with ∣x∣<1 (1/((1−x)^2 ))=Σ_(n=1) ^∞ nx^(n−1) (x/((1−x)^2 ))=Σ_(n=1) ^∞ nx^n −((1+x)/((1−x)^3 ))=Σ_(n=1) ^∞ n^2 x^(n−1) ⇔((−x(1+x))/((1−x)^3 ))=Σ_(n=1) ^∞ n^2 x^n Σ_(n=1) ^∞ n^3 x^(n−1) =(((2x+1)(1−x)+3x(1+x))/((1−x)^4 )) =((x^2 +4x+1)/((1−x)^4 )) Σ_(n=1) ^∞ n^3 x^n =((x(x^2 +4x+1))/((1−x)^4 )) x=(1/3) Σ_(n=1) ^∞ (n^3 /3^n )=(((1/9)+(4/3)+1)/(3((2/3))^4 ))=((33)/8)
$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \:\:{with}\:\:\mid{x}\mid<\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}−\mathrm{1}} \\ $$$$\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{nx}^{{n}} \\ $$$$−\frac{\mathrm{1}+{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}−\mathrm{1}} \Leftrightarrow\frac{−{x}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} {x}^{{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{3}} {x}^{{n}−\mathrm{1}} =\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}\right)+\mathrm{3}{x}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$=\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{\mathrm{3}} {x}^{{n}} =\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=\frac{\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{1}}{\mathrm{3}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{4}} }=\frac{\mathrm{33}}{\mathrm{8}} \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
correct answer thanks sir .
$${correct}\:{answer}\:{thanks}\:{sir}\:. \\ $$
Commented by Smail last updated on 02/Nov/18
Thank you for posting quistions like this.
$${Thank}\:{you}\:{for}\:{posting}\:{quistions}\:{like}\:{this}. \\ $$

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