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**3rdMath****Member**- Registered: 2013-01-29
- Posts: 1

Given the series 1/2 + 1/(2^4)+1/(2^7)+1/(2^10)

(i) show that this is geometric sequence..........can some1 help with this please

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,671

Hi;

That is a geometric series because each term has a common ratio which is

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**n872yt3r****Member**- Registered: 2013-01-21
- Posts: 392

(2^4=16) (2^7=49) (2^10=100) 1/16+1/49=0.0829081632653061224489795183673...+1/100=0.09290816326530612244897959183673...+1/2=0.59290816326530612244897959183673...

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,016

n872yt3r wrote:

(2^4=16) (2^7=49) (2^10=100) 1/16+1/49=0.0829081632653061224489795183673...+1/100=0.09290816326530612244897959183673...+1/2=0.59290816326530612244897959183673...

That is not correct. 2^7 is not 49 and 2^10 is not 100...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 105,671

Hi n872yt3r;

7^2 = 49 and 10^2 = 100

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.****No great discovery was ever made without a bold guess. **

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**mttal24****Member**- Registered: 2012-05-01
- Posts: 23

Well, to identify and prove a geometric progression the following can be used:

If

t2/t1=t3/t2=t4/t3=.....=tn/t(n-1)=r (and 'r' also represents common ratio)

then the sequence is a GP.

Here,

1/2^4 divided by1/2 is equal to 1/2^7 divided by 1/2^4.

Thus, you can show that it is a gp

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,657

hi 3rdMath

Welcome to the forum.

If you had an algebraic form for the general term, then you could do the job in one go with

As you have just 4 terms and no general term you will have to show that

The value for this constant has already been given in earlier posts.

Bob

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