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Re: Mathematical Induction
Posted:
Oct 9, 2012 5:15 PM
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The sum is over the first n terms. The math before the ellipses shows how the pattern begins and after the ellipses shows how it ends.
p(0)=2
p(1)=2-2*7
p(2)=2-2*7+2*7^2
(1-(-7)^(n+1))/4 + 2*(-7)^n=(1+(8-1)(-7)^(n+1))/4=(1-(-7)^(n+2))/4
Alternatively, prove by induction the formula for the sum of the first n term geometric progression, common ratio r and 0th term a, a(1-r^n)/(1-r). Then set a=2, r=-7.
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From: Joe <discussions@mathforum.org>
To: discretemath@mathforum.org
Sent: Tuesday, 9 October 2012, 16:12
Subject: Mathematical Induction
For the following problem:
2-2*7 + 2 * 7^2 - ? + 2* (-7)^n = (1-(-7)^n+1)/4
I understand that I need to first solve this for P(0). What I font understand is how the left side of the equation is equal to 2. Maybe I'm messing up the sequence of the equation or something but when I do the math I get:
(2-2*7 + 2 * 7^2) + (2* (-7)^0) = 88
If I ignore the first portion and just do (2* (-7)^0) I get 2. I am thinking maybe the math before the ellipses is just a example of a incrementing value of n and the math after the ellipses is the actual function p(). Is this correct?
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