## Pologirl19 Consider the infinite geometric series a. Write the first four terms of the series. b. Does the series diverge or converge? c. If the series has a sum, find the sum. one year ago one year ago

1. Pologirl19

$\sum_{x=1}^{\infty}-4\frac{ 1 }{ 3 }^{x-1}$

2. Pologirl19

Help!!

3. tkhunny

It's a geometric series with r < 1. It converges and the sum is well known. $$-4 (1 + 1/3 + 1/9 + 1/27 + ...) = \dfrac{-4}{1-(1/3)}$$ Try not to struggle over ones like this.

4. Pologirl19

wait were do you get 1/9 and 1/27. Because I got b. it converges and c. it is -6. But I have to show my work. @tkhunny

5. Pologirl19

Because I I know the formula. it is $S=\frac{ a1 }{ 1-r }$

6. tkhunny

$$\sum\limits_{x=1}^{\infty} -4\left(\dfrac{1}{3}\right)^{x-1} = -4\sum\limits_{x=1}^{\infty} \left(\dfrac{1}{3}\right)^{x-1} = -4\left(\left(\dfrac{1}{3}\right)^{1-1} + \left(\dfrac{1}{3}\right)^{2-1} + \left(\dfrac{1}{3}\right)^{3-1} + ...\right)$$ Are you yet seeing it?

7. Pologirl19

Ah thank you so much I was really confused. except for I don't get a.

8. tkhunny

(1/3)^(1-1) = (1/3)^0 = 1 -4 * 1 = -4 = a1

9. Pologirl19

Thats a.?

10. tkhunny

a1 is the first term in the series. -4 in this case.

11. Pologirl19

Wait still a little confused thats only one though

12. tkhunny

a is the first term. -4 Use x = 1 in the summation formula. r is the common ratio. 1/3 Just read it from the summation formula. $$\dfrac{-4}{1-1/3} = \dfrac{-4}{2/3}$$ You're not going to make me do ALL the work, are you?

13. Pologirl19

No sorry. So it would be -4, 1/3, 1 and infinity. Right

14. tkhunny

What would be -4? What would be 1/3? What would be 1? What would be "infinity"? Its an infinite series. Its first term is -4 Its common ration is 1/3 That is ALL the required information to find the sum.

15. Pologirl19

Well it says write the first four terms of the series

16. tkhunny

Then do that. Substitute x = 1, 2, 3, 4. All should be negative. All should be finite.