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n-1-2n-1-2-n-




Question Number 131825 by liberty last updated on 09/Feb/21
  Σ_(n=1) ^∞  ((2n−1)/2^n ) ?
n=12n12n?
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
method1n=1x2n=x21x2n=1x2n1=x1x2n=1(2n1)x2n2=11x2+2x2(1x2)2n=1(2n1)x2n=x21x2+2x4(1x2)2x2=12n=12n12n=12112+2×14(112)2=1+2=3
Commented by liberty last updated on 09/Feb/21
cool...
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
method2:n=12n12n=n=1n2n1n=112n=S1S2S2=12+(12)2+(12)3+=12112=1S1=1+221+322+423+12S1=12+222+323+424+(112)S1=1+12+122+123+=1+S2=2S1=2×2=4n=12n12n=41=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
Methodπ:)S=x+3x2+5x3+7x4+..S(1x)=x+2x2+2x2+S(1x)=2x(1x)xS=2x(1x)2x(1x)(x=12)S=41=3

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