Question

Please answer this question​

Answer 1

2

Explanation:

`\boxed{\begin{minipage}{5.5 cm}\underline{Sum of an infinite geometric series}\\\\$S_(\infty)=(a)/(1-r)$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $r$ is the common ratio.\\\end{minipage}}`

Given geometric series:

`S=(1 * 2)/(3)+(2 * 3)/(3^2)+ (3 * 4)/(3^3)+(4 * 5)/(3^4)+... \infty`

The first term a is:

`a=(1 * 2)/(3)=(2)/(3)`

The common ratio r is:

`r=(a_3)/(a_2)=( (3 * 4)/(3^3))/((2 * 3)/(3^2))=((4)/(9))/((2)/(3))=(2)/(3)`

Substitute the first term and common ratio into the formula:

`\implies S_(\infty)=((2)/(3))/(1-(2)/(3))=((2)/(3))/((1)/(3))=2`