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#1 2008-08-01 08:11:48

Sku11
Member
Registered: 2008-08-01
Posts: 3

Proof of the sum of a geometric progression PLEASE HELP

Hey guys, sorry im quite new to mathsisfun.com I just had a simple question.

This is really going to sound pathetic and its really like grade 1 maths but for the life of me I cannot figure out the proof of this formula.Here is the proof:

  Geometric Progression = a + ar + ar^2 + ar^3 + ... + ar^n-1  (1)

  multiply formula (1) by r:    r.GP = ar + ar^2 + ar^3 + ... + ar^n-1 + ar^n (2)

  subtracting formula (2) from (1)gives: GP- r.GP = a -ar^n
 
  factoring on both sides gives: GP(1-r)=a(1-r^n)

  dividing by (1-r): GP= a(1 -r^n)/ (1-r)

My problem in the proof is not the logic behind it its the mechanics of the maths in line 2 of the proof where formula 1 gets multiplied through by r, for the life of me I dont understand why by multiplying through by "r" it leaves the term "ar^n-1". I though by multiplying the term ar^n-1 in the first formula it leaves ar^n due to the laws of exponents thus I do not see where the original term comes from in the second line. Its probably one of the most fundamental laws of algebra and I feel really bad not being able to get it, I just would really like some help on it please.

P.S. I got the proof from this url http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/APGP.pdf page 9 at the bottom.
** Sorry could one of the moderators please move this thread to the Help me section**

Last edited by Sku11 (2008-08-01 08:34:45)

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#2 2008-08-01 09:21:36

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Proof of the sum of a geometric progression PLEASE HELP

Moved

The ar^n-1 in line (1) becomes the ar^n in line (2) when it gets multiplied by r.
However, there's an ar^n-2 hidden in line (1)'s "..." which becomes the ar^n-1 that you see in (2).

It might help if you think of the first line as:

GP = a + ar + ar^2 + ... + ar^n-2 + ar^n-1.
It's the same thing, I've just put different things into the "..." part.


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-08-01 10:15:49

Sku11
Member
Registered: 2008-08-01
Posts: 3

Re: Proof of the sum of a geometric progression PLEASE HELP

Ok, that clears up alot of things and makes things confusing at the same time for me tongue
I thought the term ar^n-1 was put into the equation so that the first term "a" doesnt get multiplied by r, so why is there a ar^n-2 in there? Thanks for the help by the way.

Last edited by Sku11 (2008-08-01 10:27:04)

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#4 2008-08-01 11:46:36

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Proof of the sum of a geometric progression PLEASE HELP

This is what the formula tells us:

1+10+100+1000 is 1111.
And 9999/9 is also 1111.
9999+1 is 10^4
1000 is 10^3

1+7+7*7+7*7*7=(7*7*7*7-1)/6
1+7+7*7+7*7*7= (7*7*7*7-1)/6

14+14*13+14*13*13 = 14*(13*13*13-1)/12

16+32+64+128+256 = 16*((2^5)-1)/(2-1)

ask more questions, we'll get this...

Last edited by John E. Franklin (2008-08-01 11:51:03)


igloo myrtilles fourmis

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#5 2008-08-01 11:54:32

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Proof of the sum of a geometric progression PLEASE HELP

About n and n-1 and n-2:
Say you got some numbers in a list.
4,5,6,7,8,9
Now say n=10.
You could write the list as:
n-6,n-5,n-4,n-3,n-2,n-1
and that is the same as the
4,5,6,7,8,9
is that helpful?


igloo myrtilles fourmis

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#6 2008-08-01 20:52:39

Sku11
Member
Registered: 2008-08-01
Posts: 3

Re: Proof of the sum of a geometric progression PLEASE HELP

Hmmm I understand exactly what you saying John E. Franklin I think hmm, so if the equation was expanded it would have terms ar^n-1,n-2,n-3 and so on?

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#7 2008-08-01 21:14:14

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Proof of the sum of a geometric progression PLEASE HELP

Yes, all the way down to 0. The '...' simply means that since we don't know what n is, we don't know how many terms we will need to add before reaching 0.

It's a bit like a saying I remember hearing in Primary school as a parody for having to count numbers in class: "1, 2, skip a few, 99, 100".

In this case, "0, 1, skip a few, n-2, n-1"

Last edited by Identity (2008-08-01 21:15:57)

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#8 2008-08-02 08:03:48

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Proof of the sum of a geometric progression PLEASE HELP

The "..." simply means the pattern continues.  Some examples:

1, 2, ..., 5 means 1, 2, 3, 4, 5.
1, 3, 5, ..., 11 means 1, 3, 5, 7, 9, 11.
1, 2, 4, ..., 64 means 1, 2, 4, 8, 16, 32, 64.

However, as you can see, some times this might get confusing.  So instead, the last term is written in a way that shows what the pattern is.  The last two from the above could be written:

1, 3, ..., 2(5) + 1
1, 2, 4, ..., 2^6

Even better is to write all terms in this form:

2*0 + 1, 2*1 + 1, ..., 2*5 + 1
2^0, 2^1, 2^2, ..., 2^6


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2012-02-23 18:37:21

aquaspirit39
Member
Registered: 2012-02-23
Posts: 2

Re: Proof of the sum of a geometric progression PLEASE HELP

I am sorry for reviving such an old thread, but since this is almost the same question as mine I decide to ask here instead of making a new thread.

I am exactly the opposite of Sku11, I understand the mechanics but I dont understand the reason behind this proof.

I dont understand why just by multiply the function G by r and subtract by G can produce such result? Can this be applied to other progression as well?

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#10 2012-02-23 21:42:05

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Proof of the sum of a geometric progression PLEASE HELP

hi aquaspirit39

Welcome to the forum!  smile

I'm not sure I understand what you want explained but I'll have a go anyway.  Post back if you still have problems with this.

A GP, by definition, has terms, each of which differs from the one before by being multiplied by 'r',  the common ratio.

By multiplying every term by 'r' you get back almost the same terms again.  In effect, each term just shifts position in the sum.

So, when you subtract, most of the terms cancel each other out, and you get left with a much simpler expression with only two terms.   That makes it easy to construct the formula.

Whenever I'm having trouble with algebra, I try the same with numbers.  That makes it much easier (usually) to see what's going on.

so

therefore

Can it be applied to other progressions?  That would depend on the progression.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#11 2012-02-24 03:01:16

aquaspirit39
Member
Registered: 2012-02-23
Posts: 2

Re: Proof of the sum of a geometric progression PLEASE HELP

Thank you for the explanation smile  I am aware my question is very vague, especially because English is not my mother tongue.

What I don't understand is the basis of this proof. For example, I have seen the proof by induction on geometric progression. Proof by induction use the basis that if it is true for n = 1 and we assume it is true for n = 1 to some number k and if we can show that it also work for k+1 then we can proof the validity of the geometric progression.

The proof by induction method use to proof geometric series can be applied to other progression as well by doing the same step:
1) proof n =1
2) assume correct for 1 to some number k
3) show it stands for k+1

of course they may be some progression that can't be proven this way, in that case we can just say proof by induction is not appropriate or not powerful enough to solve the problem.

I guess I can rephrase my question to: what is the general algorithm applied to the Geometric progression in order to get the proof as posted by Skul1?

Thank you for all the help smile

Last edited by aquaspirit39 (2012-02-24 03:02:21)

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#12 2012-02-24 05:51:17

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Proof of the sum of a geometric progression PLEASE HELP

hi aquaspirit39

what is the general algorithm applied to the Geometric progression in order to get the proof as posted by Skul1?

Sorry;  I cannot answer this with a name like "Proof by Induction"

I've had a look at Wikipedia and Wolfram MathWorld but have not found what you are looking for.

For me, it's just a method that may sometimes work.

A similar approach may be used for arithmetic series.  Have a look at

http://en.wikipedia.org/wiki/Arithmetic_progression

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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