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Topic: Mathematical Induction
Replies: 2   Last Post: Oct 9, 2012 8:44 PM

 Messages: [ Previous | Next ]
 Angela Richardson Posts: 42 From: UK Registered: 6/22/11
Re: Mathematical Induction
Posted: Oct 9, 2012 5:15 PM
 att1.html (1.7 K)

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.

________________________________
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?

Date Subject Author
10/9/12 Joe
10/9/12 Angela Richardson
10/9/12 Ben Brink