Question

Geometric series. In this image, the lower limit of the summation notation is “n=1”.

Answer 1

Given the infinite geometric series below

`\sum_{n\mathop{=}1}^(\infty)-4((1)/(3))^(n-1)`

a) For the first four terms, n = 1, 2, 3, 4

Where, n = 1,

`=-4((1)/(3))^(1-1)=-4((1)/(3))^0=-4(1)=-4`

Where, n = 2

`=-4((1)/(3))^(2-1)=-4((1)/(3))^1=-4((1)/(3))=-(4)/(3)`

Where, n = 3

`=-4((1)/(3))^(3-1)=-4((1)/(3))^2=-4((1)/(9))=-(4)/(9)`

Where, n = 4

`=-4((1)/(3))^(4-1)=-4((1)/(3))^3=-4((1)/(27))=-(4)/(27)`

Hence, the first four terms are

`-4,-(4)/(3),-(4)/(9),-(4)/(27)`

b) To confirm if the series converges or diverges,

`\begin{gathered} r=\frac{second\text{ term}}{first\text{ term}}=(-(4)/(3))/(-4)=-(4)/(3)*-(1)/(4)=(1)/(3) \\ r=(1)/(3) \end{gathered}`

Since, the common ratio, r is between -1 and 1,

Hence, the series converges

c) To find the sum of the series, we will apply the sum to infinity formula, which is

`\begin{gathered} S_(\infty)=(a)/(1-r) \\ Where \\ a\text{ is the first term} \\ r\text{ is the common ratio} \end{gathered}`
`\begin{gathered} S_(\infty)=(-4)/(1-(1)/(3))=(-4)/((2)/(3))=-4/(2)/(3)=-4*(3)/(2)=-6 \\ S_(\infty)=-6 \end{gathered}`

Hence, the sum is -6