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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

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} + \dots

= 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