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) ) ?

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)) ⊛

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))

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))

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) ))