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Question Number 181360 by lapache last updated on 24/Nov/22
Calcul Σ_(k=0) ^n k((1/3))^k =...
$${Calcul} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} =… \\ $$
Answered by mahdipoor last updated on 24/Nov/22
way 2: get f(x)=e^x +e^(2x) +e^(3x) +...+e^(kx) f(x)(1−e^x )=e^x −e^((k+1)x) ⇒f(x)=((e^x −e^((k+1)x) )/(1−e^x )) ⇒⇒ (df/dx)=e^x +2e^(2x) +3e^(3x) +...+ke^(kx) (df/dx)=(((e^x −(k+1)e^((k+1)x) )(1−e^x )−(−e^x )(e^x −e^((k+1)x) ))/((1−e^x )^2 ))= ⇒x=ln3 or e^x =(1/3)⇒ (df/dx)(x=ln3)=(1/3)+(2/3^2 )+....+(k/3^k ) (df/dx)(x=ln3)=((((1/3)−((k+1)/3^(k+1) ))((2/3))−(((−1)/3))((1/3)−(1/3^(k+1) )))/(((2/3))^(2 ) ))= (3/4)−((2k+3)/(4×3^k )) ⇒⇒(1/3)+(2/3^2 )+...+(k/3^k )=(3/4)−((2k+3)/(4×3^k ))
$${way}\:\mathrm{2}: \\ $$$${get}\:{f}\left({x}\right)={e}^{{x}} +{e}^{\mathrm{2}{x}} +{e}^{\mathrm{3}{x}} +…+{e}^{{kx}} \\ $$$${f}\left({x}\right)\left(\mathrm{1}−{e}^{{x}} \right)={e}^{{x}} −{e}^{\left({k}+\mathrm{1}\right){x}} \Rightarrow{f}\left({x}\right)=\frac{{e}^{{x}} −{e}^{\left({k}+\mathrm{1}\right){x}} }{\mathrm{1}−{e}^{{x}} } \\ $$$$\Rightarrow\Rightarrow \\ $$$$\frac{{df}}{{dx}}={e}^{{x}} +\mathrm{2}{e}^{\mathrm{2}{x}} +\mathrm{3}{e}^{\mathrm{3}{x}} +…+{ke}^{{kx}} \\ $$$$\frac{{df}}{{dx}}=\frac{\left({e}^{{x}} −\left({k}+\mathrm{1}\right){e}^{\left({k}+\mathrm{1}\right){x}} \right)\left(\mathrm{1}−{e}^{{x}} \right)−\left(−{e}^{{x}} \right)\left({e}^{{x}} −{e}^{\left({k}+\mathrm{1}\right){x}} \right)}{\left(\mathrm{1}−{e}^{{x}} \right)^{\mathrm{2}} }= \\ $$$$\Rightarrow{x}={ln}\mathrm{3}\:\:\:{or}\:\:\:{e}^{{x}} =\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow \\ $$$$\frac{{df}}{{dx}}\left({x}={ln}\mathrm{3}\right)=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+….+\frac{{k}}{\mathrm{3}^{{k}} } \\ $$$$\frac{{df}}{{dx}}\left({x}={ln}\mathrm{3}\right)=\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{{k}+\mathrm{1}}{\mathrm{3}^{{k}+\mathrm{1}} }\right)\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\left(\frac{−\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}^{{k}+\mathrm{1}} }\right)}{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}\:\:} }= \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}{k}+\mathrm{3}}{\mathrm{4}×\mathrm{3}^{{k}} } \\ $$$$\Rightarrow\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{{k}}{\mathrm{3}^{{k}} }=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{2}{k}+\mathrm{3}}{\mathrm{4}×\mathrm{3}^{{k}} } \\ $$
Answered by SEKRET last updated on 24/Nov/22
A= Σ_(k=0) ^n (k/3^k ) =(1/3)+(2/3^2 )+(3/3^3 )+(4/3^4 )+....+(n/3^n ) (A/3)=Σ_(k=0) ^n (k/3^(k+1) )=(1/3^2 )+(2/3^3 )+(3/3^4 )+(4/3^5 )+..+(n/3^(n+1) ) ((2A)/3) = (1/3)+(1/3^2 )+(1/3^3 )+(1/3^4 )+...+(1/3^n )+(n/3^(n+1) ) ......
$$\:\:\boldsymbol{\mathrm{A}}=\:\underset{\mathrm{k}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\boldsymbol{\mathrm{k}}}{\mathrm{3}^{\boldsymbol{\mathrm{k}}} }\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{4}} }+….+\frac{\boldsymbol{\mathrm{n}}}{\mathrm{3}^{\boldsymbol{\mathrm{n}}} } \\ $$$$\:\:\frac{\boldsymbol{\mathrm{A}}}{\mathrm{3}}=\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\frac{\boldsymbol{\mathrm{k}}}{\mathrm{3}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{4}} }+\frac{\mathrm{4}}{\mathrm{3}^{\mathrm{5}} }+..+\frac{\boldsymbol{\mathrm{n}}}{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} } \\ $$$$\:\:\frac{\mathrm{2}\boldsymbol{\mathrm{A}}}{\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{\boldsymbol{\mathrm{n}}} }+\frac{\boldsymbol{\mathrm{n}}}{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} } \\ $$$$…… \\ $$$$ \\ $$
Answered by mahdipoor last updated on 24/Nov/22
I⇒((1/3)+(2/3^2 )+(3/3^3 )+...(n/3^n ))×(1−(1/3))= (1/3)+(2/3^2 )+...+((n−1)/3^(n−1) )+(n/3^n )−((1/3^2 )+(2/3^3 )+...((n−1)/3^n )+(n/3^(n+1) ))= (1/3)+(1/3^2 )+...+(1/3^n )−(n/3^(n+1) ) II⇒((1/3)+(1/3^2 )+...+(1/3^n ))×(1−(1/3))=(1/3)−(1/3^(n+1) ) I & II ⇒Σ_(k=0) ^n k((1/3))^k =((1/3)+(2/3^2 )+...+(n/3^n ))= ((((1/3)+(2/3^2 )+...+(n/3^n ))(1−(1/3)))/((1−(1/3))))=(((1/3)+(1/3^2 )+...+(1/3^n )−(n/3^(n+1) ))/(2/3))= ((((((1/3)+(1/3^2 )+...+(1/3^n ))(1−(1/3)))/((1−(1/3))))−(n/3^(n+1) ))/(2/3))= ((((((1/3)−(1/3^(n+1) )))/(2/3))−(n/3^(n+1) ))/(2/3))=(3/4)−((3+2n)/(4×3^n ))
$${I}\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{3}^{\mathrm{3}} }+…\frac{{n}}{\mathrm{3}^{{n}} }\right)×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{{n}−\mathrm{1}}{\mathrm{3}^{{n}−\mathrm{1}} }+\frac{{n}}{\mathrm{3}^{{n}} }−\left(\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{3}} }+…\frac{{n}−\mathrm{1}}{\mathrm{3}^{{n}} }+\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} }\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }−\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} } \\ $$$${II}\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right)×\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} } \\ $$$${I}\:\&\:{II}\:\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{k}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} =\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{{n}}{\mathrm{3}^{{n}} }\right)= \\ $$$$\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{{n}}{\mathrm{3}^{{n}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }−\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} }}{\frac{\mathrm{2}}{\mathrm{3}}}= \\ $$$$\frac{\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…+\frac{\mathrm{1}}{\mathrm{3}^{{n}} }\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)}−\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} }}{\frac{\mathrm{2}}{\mathrm{3}}}= \\ $$$$\frac{\frac{\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}^{{n}+\mathrm{1}} }\right)}{\frac{\mathrm{2}}{\mathrm{3}}}−\frac{{n}}{\mathrm{3}^{{n}+\mathrm{1}} }}{\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{3}+\mathrm{2}{n}}{\mathrm{4}×\mathrm{3}^{{n}} } \\ $$

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