Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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.

$${Thank}\:{you}\:{for}\:{posting}\:{quistions}\:{like}\:{this}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com