For-a-sequence-lt-a-n-gt-a-1-2-and-a-n-1-a-n-1-3-Then-r-1-20-a-r-is-

Question Number 31952 by 24444355 last updated on 17/Mar/18

For a sequence < a_{n} >, a_{1} = 2 and

\frac{a_{n + 1}}{a_{n}} = \frac{1}{3} . Then \underset{r = 1}{\sum^{20}} a_{r} is

Commented by abdo imad last updated on 17/Mar/18

e h a v e a_{n + 1} = \frac{1}{3} a_{n} s o (a_{n}) i s a g e o m e t r i c s e q u e n c e w i t h

q = \frac{1}{3} \Rightarrow \sum_{r = 1}^{\infty} a_{r} = a_{1} \frac{1 - q^{20}}{1 - q} = 2 \frac{1 - {(\frac{1}{3})}^{20}}{1 - \frac{1}{3}}

= 2 . \frac{3}{2} (1 - \frac{1}{3^{20}}) = 3 - \frac{3}{3^{20}} = 3 - \frac{1}{3^{19}} .