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Ben Brink
Posts:
201
From:
Rosenberg, TX
Registered:
11/11/06


RE: Mathematical Induction
Posted:
Oct 9, 2012 8:44 PM



Joe, The "n" at the end is the final integer value, but it starts with 0. In fact, if n = 0 the left side ends with 2*(7)^0 = 2*1 = 2, while the right side is just (1(7)^(0+1))/4 = (1(7))/4 = 8/4 = 2. Therefore the equation "works" for n = 0. The harder part is showing that if it works with n replaced by k, then it also works for n replaced by k+1. Good luck, and come back with any further questions. Ben
> Date: Tue, 9 Oct 2012 11:12:30 0400 > From: discussions@mathforum.org > To: discretemath@mathforum.org > Subject: Mathematical Induction > > For the following problem: > > 22*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: > > > (22*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?



