Given b_n = 3.2^n is a GP . find the value of (1/b_1 )+(1/b_2 )+(1/b_3 )+...+(1/b


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Question Number 103670 by bobhans last updated on 16/Jul/20

$G i v e n b_{n} = 3 . 2^{n} i s a G P . f i n d t h e v a l u e$ $o f \frac{1}{b_{1}} + \frac{1}{b_{2}} + \frac{1}{b_{3}} + . . . + \frac{1}{b_{10}} ?$
	Answered by bramlex last updated on 16/Jul/20
	$(1/(3.2)) + (1/(3.4))+(1/(3.8))+...+(1/(3.2^(10) )) = (1/3){(1/2)+(1/4)+(1/8)+...+(1/2^(10) )} = (1/3){(1/2). ((((1−((1/2))^(10) )/(1/2)))}= (1/3).{((2^(10) −1)/2^(10) )} = ((1023)/(3×1024))=((341)/(1024)) ⊛$
	$\frac{1}{3.2} + \frac{1}{3.4} + \frac{1}{3.8} + . . . + \frac{1}{3 . 2^{10}} =$ $\frac{1}{3} {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . + \frac{1}{2^{10}}} =$ $\frac{1}{3} {\frac{1}{2} . (\frac{(1 - {(\frac{1}{2})}^{10}}{\frac{1}{2}})} =$ $\frac{1}{3} . {\frac{2^{10} - 1}{2^{10}}} = \frac{1023}{3 \times 1024} = \frac{341}{1024} ⊛$
	Answered by Worm_Tail last updated on 16/Jul/20

	$b_{n} = 3 . 2^{n} \Rightarrow \frac{1}{b_{n}} = \frac{1}{3 {(2)}^{n}}$ $Σ \frac{1}{b_{n}} = Σ \frac{1}{3 {(2)}^{n}} = \frac{1}{3} Σ {(\frac{1}{2})}^{n}$ $Σ \frac{1}{b_{n}} = \frac{1}{3} Σ {(\frac{1}{2})}^{n} = \frac{1}{6} (\frac{1 - 2^{- n}}{0.5})$ $Σ \frac{1}{b_{n}} = \frac{1}{3} (1 - 2^{- n}) n = 10$ $Σ \frac{1}{b_{n}} = \frac{1}{3} (1 - 2^{- 10}) = \frac{341}{1024}$
	Answered by Dwaipayan Shikari last updated on 16/Jul/20

	$\frac{1}{3.2} + \frac{1}{3 . 2^{2}} + \frac{1}{3 . 2^{3}} + . . . + n = \frac{1}{3} . \frac{1}{2} (\frac{{(\frac{1}{2})}^{n} - 1}{\frac{1}{2} - 1}) = \frac{1}{3} (1 - 2^{- n}) = \frac{1}{3} . \frac{1023}{1024} = \frac{341}{1024}$
	Answered by mathmax by abdo last updated on 16/Jul/20
	$Σ_(k=1) ^(10) (1/b_k ) =Σ_(k=1) ^(10) (1/(3.2^k )) =(1/6)Σ_(k=1) ^(10) (1/2^(k−1) ) =(1/6)Σ_(k=0) ^9 ((1/2))^k =(1/6)×((1−((1/2))^(10) )/(1−(1/2))) =(1/3){1−(1/2^(10) )} =(1/3)−(1/(3.2^(10) ))$
	$\sum_{k = 1}^{10} \frac{1}{b_{k}} = \sum_{k = 1}^{10} \frac{1}{3 . 2^{k}} = \frac{1}{6} \sum_{k = 1}^{10} \frac{1}{2^{k - 1}} = \frac{1}{6} \sum_{k = 0}^{9} {(\frac{1}{2})}^{k}$ $= \frac{1}{6} \times \frac{1 - {(\frac{1}{2})}^{10}}{1 - \frac{1}{2}} = \frac{1}{3} {1 - \frac{1}{2^{10}}} = \frac{1}{3} - \frac{1}{3 . 2^{10}}$
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