Question

Please help right away.

Answer 1

$\sum\limits_{k=1}^{\infty}420\left(\dfrac{1}{6}\right)^{k-1}$

The infinite geometric series is converges if |r| < 1.

We have r=1/6 < 1, therefore our infinite geometric series is converges.

The sum S of an infinite geometric series with |r| < 1 is given by the formula :

$S=\dfrac{a_1}{1-r}$

We have:

$a_1=420\left(\dfrac{1}{6}\right)^{1-1}=420\left(\dfrac{1}{6}\right)^0=420\\\\r=\dfrac{1}{6}$

substitute:

$S=\dfrac{420}{1-\frac{1}{6}}=\dfrac{420}{\frac{5}{6}}=420\cdot\dfrac{6}{5}=84\cdot6=504$

Answer: d. Converges, 504.