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Question Number 192149 by Shrinava last updated on 09/May/23

Find: (1/2) + (3/2^3 ) + (5/2^5 ) + (7/2^7 ) + ...

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:+\:\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }\:+\:... \\ $$

Answered by aleks041103 last updated on 09/May/23

Σ_(k=1) ^∞ (2k−1)x^(2k−1) =x(Σ_(k=1) ^∞ x^(2k−1) )′= =x(xΣ_(k=1) ^∞ x^(2(k−1)) )′=x(x(1/(1−x^2 )))′= =x(d/dx)((x/(1−x^2 )))=x(((1−x^2 )−(−2x)x)/(1−x^2 ))= =x((1+x^2 )/(1−x^2 )) ⇒Σ_(k=1) ^∞ ((2k−1)/2^(2k−1) )=(x((1+x^2 )/(1−x^2 )))_(x=1/2) =(1/2) ((1+(1/4))/(1−(1/4)))= =(1/2) (5/3) ⇒(1/2)+(3/2^3 )+(5/2^5 )+...=(5/6)

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right){x}^{\mathrm{2}{k}−\mathrm{1}} ={x}\left(\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{k}−\mathrm{1}} \right)'= \\ $$$$={x}\left({x}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}\left({k}−\mathrm{1}\right)} \right)'={x}\left({x}\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)'= \\ $$$$={x}\frac{{d}}{{dx}}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)={x}\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\left(−\mathrm{2}{x}\right){x}}{\mathrm{1}−{x}^{\mathrm{2}} }= \\ $$$$={x}\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}} }=\left({x}\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\right)_{{x}=\mathrm{1}/\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+...=\frac{\mathrm{5}}{\mathrm{6}} \\ $$

Commented by deleteduser1 last updated on 09/May/23

(d/dx)((x/(1−x^2 )))=((1+x^2 )/((+x^2 −1)^2 )) ⇒x(d/dx)((x/(1−x^2 )))=((x(1+x^2 ))/((+x^2 −1)^2 )) ⇒For x=(1/2); we get ((10)/9)

$$\frac{{d}}{{dx}}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(+{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}\frac{{d}}{{dx}}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(+{x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{For}\:{x}=\frac{\mathrm{1}}{\mathrm{2}};\:{we}\:{get}\:\frac{\mathrm{10}}{\mathrm{9}} \\ $$

Commented by BaliramKumar last updated on 09/May/23

(1/2) + (3/2^3 ) + (5/2^5 ) = ((33)/(32)) > 1

$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:=\:\frac{\mathrm{33}}{\mathrm{32}}\:>\:\mathrm{1} \\ $$

Answered by mehdee42 last updated on 09/May/23

we will consider ; A=x+3x^3 +5x^5 +... ; 0<x<1 ⇒x^2 A=x^3 +3x^5 +5x^7 +... ⇒A−Ax^2 =x+2(x^3 +x^5 +x^7 +...) (1−x^2 )A=x+((2x^3 )/(1−x^2 ))⇒A=((x+x^3 )/((1−x^2 )^2 )) now regarding the question , if “x=(1/2) ” we will have answer = ((10)/9) ✓

$${we}\:{will}\:{consider}\:;\:{A}={x}+\mathrm{3}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{5}} +...\:\:\:\:;\:\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {A}={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{5}} +\mathrm{5}{x}^{\mathrm{7}} +... \\ $$$$\Rightarrow{A}−{Ax}^{\mathrm{2}} ={x}+\mathrm{2}\left({x}^{\mathrm{3}} +{x}^{\mathrm{5}} +{x}^{\mathrm{7}} +...\right) \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){A}={x}+\frac{\mathrm{2}{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{2}} }\Rightarrow{A}=\frac{{x}+{x}^{\mathrm{3}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${now}\:{regarding}\:\:{the}\:{question}\:,\:{if}\:\:``{x}=\frac{\mathrm{1}}{\mathrm{2}}\:''\:{we}\:{will}\:{have} \\ $$$${answer}\:=\:\frac{\mathrm{10}}{\mathrm{9}}\:\checkmark \\ $$$$ \\ $$

Answered by universe last updated on 09/May/23

s = (1/2)+(3/2^3 )+(5/2^5 )+(7/2^7 )+... (1/4)s = (1/2^3 )+(3/2^5 )+(5/2^7 )+... (3/4)s = (1/2)+(1/2^2 )+(1/2^4 )+(1/2^6 )+... (3/4)s = (1/2) + ((1/2^2 )/(1−1/2^2 )) (3/4)s = (1/2) + (1/3) s = (5/6)×(4/3) = ((20)/(18)) s = ((10)/9)

$$\:\mathrm{s}\:\:\:\:\:\:\:=\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }+... \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{s}\:\:\:=\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{7}} }+... \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\mathrm{s}\:\:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }+... \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}\mathrm{s}\:\:\:=\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:+\:\:\frac{\mathrm{1}/\mathrm{2}^{\mathrm{2}} }{\mathrm{1}−\mathrm{1}/\mathrm{2}^{\mathrm{2}} } \\ $$$$\:\frac{\mathrm{3}}{\mathrm{4}}\mathrm{s}\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{s}\:=\:\:\frac{\mathrm{5}}{\mathrm{6}}×\frac{\mathrm{4}}{\mathrm{3}}\:=\:\frac{\mathrm{20}}{\mathrm{18}} \\ $$$$\mathrm{s}\:=\:\frac{\mathrm{10}}{\mathrm{9}} \\ $$

Answered by York12 last updated on 09/May/23

that′s AGP

$${that}'{s}\:{AGP} \\ $$

Answered by manxsol last updated on 11/May/23

t_k =q ((2k−1)/2^(2k−1 ) )

$$\:{t}_{{k}} ={q}\:\:\:\:\:\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}^{\mathrm{2}{k}−\mathrm{1}\:\:} } \\ $$

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