Σ_(r=0) ^∞ 2^(4r−2)


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Question Number 36837 by Rio Mike last updated on 06/Jun/18

$\underset{r = 0}{\sum^{\infty}} 2^{4 r - 2}$
Commented by prof Abdo imad last updated on 06/Jun/18
$let S_n =Σ_(k=0) ^n 2^(4k −2) ⇒S_n =(1/4) Σ_(k=0) ^n (2^4 )^k =(1/4) (((2^4 )^(n+1) −1)/(2^4 −1)) = (1/(4(2^4 −1))){ 2^(4n+4) −1} but 2>1 ⇒ lim_(n→+∞) 2^(4n+4) =+∞ ⇒lim_(n→+∞) S_n =+∞$
$l e t S_{n} = \sum_{k = 0}^{n} 2^{4 k - 2} \Rightarrow S_{n} = \frac{1}{4} \sum_{k = 0}^{n} {(2^{4})}^{k}$ $= \frac{1}{4} \frac{{(2^{4})}^{n + 1} - 1}{2^{4} - 1} = \frac{1}{4 (2^{4} - 1)} {2^{4 n + 4} - 1} b u t$ $2 > 1 \Rightarrow {l i m}_{n \to + \infty} 2^{4 n + 4} = + \infty \Rightarrow {l i m}_{n \to + \infty} S_{n} = + \infty$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18
	$s=2^(4.0−2) +2^(4.1−2) +2^(4.2−2) +2^(4.3−2) +... =2^(−2) +2^2 +2^6 +2^(10) +.. =((2^(−2) {(2^2 )^n −1})/(2^2 −1)) =((2^(−2) (2^(2n) −1))/3)=((2^(2n) −1)/(12)) since common ratio is 2^2 i,e 4>1 so s→∞$
	$s = 2^{4.0 - 2} + 2^{4.1 - 2} + 2^{4.2 - 2} + 2^{4.3 - 2} + . . .$ $= 2^{- 2} + 2^{2} + 2^{6} + 2^{10} + . .$ $= \frac{2^{- 2} {{(2^{2})}^{n} - 1}}{2^{2} - 1}$ $= \frac{2^{- 2} (2^{2 n} - 1)}{3} = \frac{2^{2 n} - 1}{12}$ $s i n c e c o m m o n r a t i o i s 2^{2} i, e 4 > 1$ $s o s \to \infty$
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