SOLUTION: Hello! I’m really stumped on this. The question says to consider the infinite geometric series:
sigma infinity n=1 -4(1/3)^n-1
and it says the lower limit of the summation
Algebra ->
Sequences-and-series
-> SOLUTION: Hello! I’m really stumped on this. The question says to consider the infinite geometric series:
sigma infinity n=1 -4(1/3)^n-1
and it says the lower limit of the summation
Log On
Question 1141362: Hello! I’m really stumped on this. The question says to consider the infinite geometric series:
sigma infinity n=1 -4(1/3)^n-1
and it says the lower limit of the summation notation is “n=1”. It then says to write the first four terms of the series, asks if the series diverge or converge, and if the series has a sum, find it. I think the series converge, and if I solved for the sum correctly, I got -5.9..., but I can’t figure out the first four terms right. Can you help? And have I solved the rest correctly? Answer by math_helper(2461) (Show Source):
then your answer for the sum of the first four terms is correct (more precisely, the answer is -160/27 which is about -5.925925...).
The sum of n terms can be found from this:
where r is the common ratio (1/3) and is the first term (-4) ( so you do need to plug in n=1 to find )
Example:
You could also plug in n=1,2,3,4 into the summation and write it all out long-hand but its a lot of extra work.
As n-->, the term goes to zero, leaving just as the sum of the infinite progression:
