Question

use the formula to evaluate the series 6+2+2/3+2/9

Answer 1

If you're talking about adding just these four terms, notice that

$S_4=6+2+\dfrac23+\dfrac29$
$S_4=6\left(1+\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}\right)$
$\dfrac13S_4=6\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\dfrac1{3^4}\right)$

$\implies S_4-\dfrac13S_4=6\left(1-\dfrac1{3^4}\right) [tex]\dfrac23S_4=6\left(1-\dfrac1{81}\right)$
$S_4=9\times\dfrac{80}{81}$
$S_4=\dfrac{80}9$

If you were referring to an infinite sum, so that the pattern continues, note that the same procedure as above can be applied to the $n$th partial sum:

$S_n=6\left(1+\dfrac13+\dfrac1{3^2}+\cdots+\dfrac1{3^n}\right)$
$\dfrac23S_n=6\left(1-\dfrac1{3^{n+1}}\right)$
$S_n=9\left(1-\dfrac1{3^{n+1}}\right)$

Then as $n\to\infty$, the exponential term approaches 0, leaving you with

$6+2+\dfrac23+\dfrac29+\cdots=9$