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Question Number 131825 by liberty last updated on 09/Feb/21
  Σ_(n=1) ^∞  ((2n−1)/2^n ) ?
$$\:\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2n}−\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:? \\ $$
Answered by mr W last updated on 09/Feb/21
method 1  Σ_(n=1) ^∞ x^(2n) =(x^2 /(1−x^2 ))  Σ_(n=1) ^∞ x^(2n−1) =(x/(1−x^2 ))  Σ_(n=1) ^∞ (2n−1)x^(2n−2) =(1/(1−x^2 ))+((2x^2 )/((1−x^2 )^2 ))  Σ_(n=1) ^∞ (2n−1)x^(2n) =(x^2 /(1−x^2 ))+((2x^4 )/((1−x^2 )^2 ))  x^2 =(1/2)  Σ_(n=1) ^∞ ((2n−1)/2^n )=((1/2)/(1−(1/2)))+((2×(1/4))/((1−(1/2))^2 ))=1+2=3
$${method}\:\mathrm{1} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} =\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}−\mathrm{1}} =\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{\mathrm{2}{n}−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{n}−\mathrm{1}\right){x}^{\mathrm{2}{n}} =\frac{{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }+\frac{\mathrm{2}{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}^{{n}} }=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}}{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$
Commented by liberty last updated on 09/Feb/21
cool...
$$\mathrm{cool}… \\ $$
Commented by mr W last updated on 09/Feb/21
method 2:  Σ_(n=1) ^∞ ((2n−1)/2^n )  =Σ_(n=1) ^∞ (n/2^(n−1) )−Σ_(n=1) ^∞ (1/2^n )=S_1 −S_2   S_2 =(1/2)+((1/2))^2 +((1/2))^3 +...=((1/2)/(1−(1/2)))=1  S_1 =1+(2/2^1 )+(3/2^2 )+(4/2^3 )+...  (1/2)S_1 =(1/2)+(2/2^2 )+(3/2^3 )+(4/2^4 )+...  (1−(1/2))S_1 =1+(1/2)+(1/2^2 )+(1/2^3 )+...=1+S_2 =2  ⇒S_1 =2×2=4  ⇒Σ_(n=1) ^∞ ((2n−1)/2^n )=4−1=3
$${method}\:\mathrm{2}: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}^{{n}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{\mathrm{2}^{{n}−\mathrm{1}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{n}} }={S}_{\mathrm{1}} −{S}_{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}} +…=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{1} \\ $$$${S}_{\mathrm{1}} =\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{1}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{3}} }+… \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{S}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{4}} }+… \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right){S}_{\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+…=\mathrm{1}+{S}_{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{S}_{\mathrm{1}} =\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}^{{n}} }=\mathrm{4}−\mathrm{1}=\mathrm{3} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Feb/21
Method ⌊π⌋:)               S=x+3x^2 +5x^3 +7x^4 +..  S(1−x)=x+2x^2 +2x^2 +...  S(1−x)=((2x)/((1−x)))−x⇒S=((2x)/((1−x)^2 ))−(x/((1−x)))    (x=(1/2))  S=4−1=3
$$\left.{Method}\:\lfloor\pi\rfloor:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{S}={x}+\mathrm{3}{x}^{\mathrm{2}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{4}} +.. \\ $$$${S}\left(\mathrm{1}−{x}\right)={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} +… \\ $$$${S}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)}−{x}\Rightarrow{S}=\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }−\frac{{x}}{\left(\mathrm{1}−{x}\right)}\:\:\:\:\left({x}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${S}=\mathrm{4}−\mathrm{1}=\mathrm{3} \\ $$

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