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Question 872382: 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.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Geometric Progression = GP =
a + ar + ar^2 + ar^3 + ... + ar^(n-2) + ar^n-1 (1)
multiply formula (1) by r to get:
r.GP = ar + ar^2 + ar^3 + ... + ar^(n-1) + ar^n (2)
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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)
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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".
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Ans: ar^(n-1) is the result of multiplying ar^(n-2) by r
r[ar^(n-2) = ar^(n-2+1) = ar^(n-1)
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Each term of (2) simply adds "1" to the exponent of "r"
in (1).
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Cheers,
Stan H.
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