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Infinite geometric series

Hi,
I have a question I'm stuck on, I have been looking for help online but can't find what I'm looking for.

I have already worked out that my first term (a) = 96 and my common ratio (r) = 0.4 in part a) of my question.
In part b) of my question I worked out the sum to infinity of the series using a1 / 1 - r , which was 160.

For part c I have been asked to find the value of

infinity
(sigma notation) Un
n=4

I am confused with what I need to do here as n = 4.

Hope this is easy to understand, thank you in advance :smile:
Original post by ubunu0
Hi,
I have a question I'm stuck on, I have been looking for help online but can't find what I'm looking for.

I have already worked out that my first term (a) = 96 and my common ratio (r) = 0.4 in part a) of my question.
In part b) of my question I worked out the sum to infinity of the series using a1 / 1 - r , which was 160.

For part c I have been asked to find the value of

infinity
(sigma notation) Un
n=4

I am confused with what I need to do here as n = 4.

Hope this is easy to understand, thank you in advance :smile:

The series sum from n = 4 to n = infinity will be the same as (the sum from n = 1 to n = infinity) minus (the sum from n = 1 to n = 3).
Reply 2
Could you possibly post a picture of the question?
Original post by ubunu0
Hi,
I have a question I'm stuck on, I have been looking for help online but can't find what I'm looking for.

I have already worked out that my first term (a) = 96 and my common ratio (r) = 0.4 in part a) of my question.
In part b) of my question I worked out the sum to infinity of the series using a1 / 1 - r , which was 160.

For part c I have been asked to find the value of

infinity
(sigma notation) Un
n=4

I am confused with what I need to do here as n = 4.

Hope this is easy to understand, thank you in advance :smile:

Actually not very easy to understand - it's generally better to post the whole question. In terms of layout: if you don't know how to use LaTeX (which lets you actually type equations like n=4Un\sum_{n=4}^\infty U_n in the forum), it's better to actually use words: "The sum from n = 4 to infinity of U_n".
Original post by old_engineer
The series sum from n = 4 to n = infinity will be the same as (the sum from n = 1 to n = infinity) minus (the sum from n = 1 to n = 3).

True (and the simplest to understand).

For long term understanding, it's probably better to write out the first few terms of both n=0arn\sum_{n=0}^\infty a r^n and n=4arn\sum_{n=4}^\infty ar^n and deduce how one is a (simple) multiple of the other.
Reply 4
Original post by ram8_
Could you possibly post a picture of the question?

Screenshot 2020-12-12 at 10.03.01.png
Reply 5
Original post by DFranklin
Actually not very easy to understand - it's generally better to post the whole question. In terms of layout: if you don't know how to use LaTeX (which lets you actually type equations like n=4Un\sum_{n=4}^\infty U_n in the forum), it's better to actually use words: "The sum from n = 4 to infinity of U_n".

True (and the simplest to understand).

For long term understanding, it's probably better to write out the first few terms of both n=0arn\sum_{n=0}^\infty a r^n and n=4arn\sum_{n=4}^\infty ar^n and deduce how one is a (simple) multiple of the other.

I will look into using LaTeX if I have future queries - I was not aware of it. Sorry!
Reply 6
Original post by ubunu0
I will look into using LaTeX if I have future queries - I was not aware of it. Sorry!

Note you could simply use the normal sum from 1..infinity with initial term
ar^3
Saves having to subtract off the first three terms.
Reply 7
Original post by mqb2766
Note you could simply use the normal sum from 1..infinity with initial term
ar^3
Saves having to subtract off the first three terms.

Thank you for that! Confirmed my answer :smile:

Appreciate all the swift responses, have a lovely weekend everybody

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