Question

Please answer this question​

Answer 1

Answer:

2

Step-by-step explanation:

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

Given geometric series:

$S=\dfrac{1 \times 2}{3}+\dfrac{2 \times 3}{3^2}+ \dfrac{3 \times 4}{3^3}+\dfrac{4 \times 5}{3^4}+... \infty$

The first term a is:

$a=\dfrac{1 \times 2}{3}=\dfrac{2}{3}$

The common ratio r is:

$r=\dfrac{a_3}{a_2}=\dfrac{ \dfrac{3 \times 4}{3^3}}{\dfrac{2 \times 3}{3^2}}=\dfrac{\dfrac{4}{9}}{\dfrac{2}{3}}=\dfrac{2}{3}$

Substitute the first term and common ratio into the formula:

$\implies S_{\infty}=\dfrac{\dfrac{2}{3}}{1-\dfrac{2}{3}}=\dfrac{\dfrac{2}{3}}{\dfrac{1}{3}}=2$