Question

How do I calculate the sum of the infinite series:

(4/9)^3 - (4/9)^4 + (4/9)^5 - (4/9)^6 + ...

Giving the answer as a fraction?

Answer 1

Chris do you know some LaTeX reader ? I wanted to practice LaTeX.
I tried MikTex but doesn't work on my PC.

Answer 2

For the question anaria take - terms and plus terms then sum (G.P sum) and then normal subtraction

Answer 3

sum of an infinite geometric series is:

S=a/(1-r)

= (4/9)^3/(1-4/9)
=64/405
=

Answer 4

raheen ... thanks. looks right to me. :)

Answer 5

geometric series
a1 = (4/9)^3
r = -4/9

\[s = \frac{a_{1}}{1-r} = \frac{(4/9)^{3}}{1+4/9} = \frac{4^{3}}{9^{3}}*\frac{9}{13} = \frac{64}{81*13} \approx 0.06\]

Answer 6

ishaan that would be ok if not an infinite sequence

Answer 7

wish that anaria

Answer 8

yeah your right chris and raheen did that correct

Answer 9

r should be negative, making denominator (1+4/9) making the sum less

Answer 10

Ishaan ... I am not Chris. I think you must have me confused with someone else.

Confusingly, dumbcow also has an option 64/1053 which is also one of the options! Now which one is right. lol

Answer 11

chris (dumbcow).........he did that right

Answer 12

If a series in alternating sign, then the common ratio must be negative :)

Answer 13

Ok guys thanks ... :)