Hello, I need help with the compute and giving exact values

Question

asked Sep 4, 2023 126k views

1 Answer

Chofoteddy · Answer 1 · 2023-09-10T10:26:50+0000

Notice that in both cases we have geometric series. Then, we have the following general rule:

$\begin{gathered} \Sigma_(k=0)^(\infty)(ar^k)=(a)/(1-r) \\ if\text{ \mid r\mid<1} \end{gathered}$

then, in the first case we have the following sum:

$\Sigma_(k=0)^(\infty)3(-(1)/(5))^k=(3)/(1-(-(1)/(5)))=(3)/(1+(1)/(5))=(3)/((6)/(5))=(15)/(6)=(5)/(2)$

for the second case, we hve:

$\Sigma_(k=1)^(\infty)2(-(3)/(5))^k=(2(-(3)/(5)))/(1-(-(3)/(5)))=(-(6)/(5))/(1+(3)/(5))=(-(6)/(5))/((8)/(5))=(-30)/(40)=-(3)/(4)$