Question

Consider the infinite geometric series infinity sigma n=1 -4(2/3)^n-1. In this, the lower limit of the summation notion is "n=1". PLEASE show your work I dont understand

1)write the first four terms of the series 2) does the series diverge or coverge 3) if the series has a sum find the sum

Answer 1

To find the first four terms of the series, we can substitute the values of n from 1 to 4 in the given formula and simplify:

n=1: -4(2/3)^0 = -4(1) = -4

n=2: -4(2/3)^1 = -4(2/3) = -8/3

n=3: -4(2/3)^2 = -4(4/9) = -16/9

n=4: -4(2/3)^3 = -4(8/27) = -32/27

So the first four terms of the series are -4, -8/3, -16/9, -32/27.

To determine if the series converges or diverges, we need to find the common ratio r, which is the ratio of any term to its preceding term:

r = (-4(2/3)^n) / (-4(2/3)^(n-1)) = 2/3

Since the absolute value of the common ratio is less than 1, the series converges.

To find the sum of the series, we can use the formula for the sum of an infinite geometric series:

S = a / (1 - r)

where a is the first term and r is the common ratio. Substituting the values we found:

S = -4 / (1 - 2/3) = -4 / (1/3) = -12

So the sum of the infinite geometric series is -12.

Answer 2

a) The first four terms of the series are:

-4, -8/3, -16/9, -32/27

b) The series converges.

c) The sum of the infinite geometric series is -12.

Explanation:

Given infinite geometric series:

`\displaystyle \sum^(\infty)_(n=1) -4 \left((2)/(3)\right)^(n-1)`

Therefore, the formula for the nth term of the sequence is:

`a_n=-4 \left((2)/(3)\right)^(n-1)`

where:

• -4 is the first term, a.
• 2/3 is the common ratio, r.

Calculate the first four terms of the series by substituting the n-values 1 through 4 into the nth term formula.

`a_1 = -4 \left((2)/(3)\right)^(1-1) = -4 \left((2)/(3)\right)^(0)=-4`

`a_2 = -4 \left((2)/(3)\right)^(2-1)= -4 \left((2)/(3)\right)^(1)=-(8)/(3)`

`a_3 = -4 \left((2)/(3)\right)^(3-1)= -4 \left((2)/(3)\right)^(2)=-4 \left((4)/(9)\right)=-(16)/(9)`

`a_4 = -4 \left((2)/(3)\right)^(4-1)= -4 \left((2)/(3)\right)^(3)= -4 \left((8)/(27)\right)=-(32)/(27)`

A geometric series converges if the absolute value of its common ratio is less than 1. As r = 2/3 and |2/3| < 1, the series converges.

Use the infinite geometric series sum formula to find sum of the infinite series.

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

Substitute a = -4 and r = 2/3 into the formula to find the sum of the infinite series:

`\begin{aligned}\implies S_(\infty)&=(-4)/(1-(2)/(3))\\\\ &=(-4)/((1)/(3))\\\\&=-4 \cdot (3)/(1)\\\\&=-(12)/(1)\\\\&=-12\end{aligned}`

Therefore, the sum of the given infinite geometric series is -12.