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Given-b-n-3-2-n-is-a-GP-find-the-value-of-1-b-1-1-b-2-1-b-3-1-b-10-

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Question Number 103670 by bobhans last updated on 16/Jul/20
Given b_n = 3.2^n is a GP . find the value of (1/b_1 )+(1/b_2 )+(1/b_3 )+...+(1/b_(10) ) ?
$${Given}\:{b}_{{n}} \:=\:\mathrm{3}.\mathrm{2}^{{n}} \:{is}\:{a}\:{GP}\:.\:{find}\:{the}\:{value} \\ $$$${of}\:\frac{\mathrm{1}}{{b}_{\mathrm{1}} }+\frac{\mathrm{1}}{{b}_{\mathrm{2}} }+\frac{\mathrm{1}}{{b}_{\mathrm{3}} }+…+\frac{\mathrm{1}}{{b}_{\mathrm{10}} }\:?\: \\ $$
Answered by bramlex last updated on 16/Jul/20
(1/(3.2)) + (1/(3.4))+(1/(3.8))+...+(1/(3.2^(10) )) = (1/3){(1/2)+(1/4)+(1/8)+...+(1/2^(10) )} = (1/3){(1/2). ((((1−((1/2))^(10) )/(1/2)))}= (1/3).{((2^(10) −1)/2^(10) )} = ((1023)/(3×1024))=((341)/(1024)) ⊛
$$\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{8}}+…+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}^{\mathrm{10}} }\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+…+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\right\}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{1}}{\mathrm{2}}.\:\left(\frac{\left(\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{10}} \right.}{\frac{\mathrm{1}}{\mathrm{2}}}\right)\right\}= \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}.\left\{\frac{\mathrm{2}^{\mathrm{10}} −\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\right\}\:=\:\frac{\mathrm{1023}}{\mathrm{3}×\mathrm{1024}}=\frac{\mathrm{341}}{\mathrm{1024}}\:\circledast \\ $$
Answered by Worm_Tail last updated on 16/Jul/20
b_n =3.2^n ⇒(1/b_n )=(1/(3(2)^n )) Σ(1/b_n )=Σ(1/(3(2)^n ))=(1/3)Σ((1/2))^n Σ(1/b_n )=(1/3)Σ((1/2))^n =(1/6)(((1−2^(−n) )/(0.5))) Σ(1/b_n )=(1/3)(1−2^(−n) ) n=10 Σ(1/b_n )=(1/3)(1−2^(−10) ) =((341)/(1024))
$${b}_{{n}} =\mathrm{3}.\mathrm{2}^{{n}} \Rightarrow\frac{\mathrm{1}}{{b}_{{n}} }=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}\right)^{{n}} } \\ $$$$\:\:\:\:\:\Sigma\frac{\mathrm{1}}{{b}_{{n}} }=\Sigma\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}\right)^{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$$$\:\:\:\:\:\Sigma\frac{\mathrm{1}}{{b}_{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}\Sigma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}−\mathrm{2}^{−{n}} }{\mathrm{0}.\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\Sigma\frac{\mathrm{1}}{{b}_{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}^{−{n}} \right)\:\:\:\:\:{n}=\mathrm{10} \\ $$$$\:\:\:\:\:\Sigma\frac{\mathrm{1}}{{b}_{{n}} }=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}^{−\mathrm{10}} \right)\:\:\:\:=\frac{\mathrm{341}}{\mathrm{1024}} \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 16/Jul/20
(1/(3.2))+(1/(3.2^2 ))+(1/(3.2^3 ))+...+n=(1/3).(1/2)(((((1/2))^n −1)/((1/2)−1)))=(1/3)(1−2^(−n) )=(1/3).((1023)/(1024))=((341)/(1024))
$$\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}^{\mathrm{3}} }+…+{n}=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} −\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{2}^{−{n}} \right)=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1023}}{\mathrm{1024}}=\frac{\mathrm{341}}{\mathrm{1024}} \\ $$
Answered by mathmax by abdo last updated on 16/Jul/20
Σ_(k=1) ^(10) (1/b_k ) =Σ_(k=1) ^(10) (1/(3.2^k )) =(1/6)Σ_(k=1) ^(10) (1/2^(k−1) ) =(1/6)Σ_(k=0) ^9 ((1/2))^k =(1/6)×((1−((1/2))^(10) )/(1−(1/2))) =(1/3){1−(1/2^(10) )} =(1/3)−(1/(3.2^(10) ))
$$\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\mathrm{b}_{\mathrm{k}} }\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}^{\mathrm{k}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{10}} \:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{k}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{9}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}×\frac{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{10}} }{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{10}} }\right\}\:=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{2}^{\mathrm{10}} } \\ $$

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