Mathematical Induction Question

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ScripterKitty
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Mathematical Induction Question

Postby ScripterKitty » Tue Oct 09, 2012 2:40 pm

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 dont 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 elipses is just a example of a incrementing value of n and the math after the elipses is the actual function p(). Is this correct?

anonmeans
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Re: Mathematical Induction Question

Postby anonmeans » Tue Oct 09, 2012 5:55 pm

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 do you mean by "solving"? Aren't you supposed to do the base step with n = 0 or n = 1 or whatever start value they gave you?
What I dont 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
What is the formula for the terms? It looks like it's 2*7^n with the start value as n = 0. If so, how are you getting the terms you listed? What was your reasoning?

ScripterKitty
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Re: Mathematical Induction Question

Postby ScripterKitty » Tue Oct 09, 2012 7:21 pm

The problem:

2-2*7 + 2 * 7^2 - … + 2* (-7)^n = (1-(-7)^n+1)/4

is to be proved P(n). I already know the answer to the problem. What I am trying to do is figure out how it was found. Here is a snipped for the answer:

"In order to prove this for all integers n>= 0, we first prove the base case P(0) and then prove the inductive step, that P(n) implies P(n+1). Now in P(0), the left-hand side has just one term, namely 2, "

What I dont understand is how was the left side of the equation narrowed downed to only 1 term which was 2. Appreciate if anyone can help explain.

anonmeans
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Re: Mathematical Induction Question

Postby anonmeans » Tue Oct 09, 2012 7:42 pm

2-2*7 + 2 * 7^2 - … + 2* (-7)^n = (1-(-7)^n+1)/4 is to be proved P(n).

"In order to prove this for all integers n>= 0....
Right there is more information: You're supposed to prove (not "solve") the statement "for all n greater than or equal to zero." I'm guessing "proving P(n)" means "proving by induction"??
"we first prove the base case P(0) and then prove the inductive step, that P(n) implies P(n+1). Now in P(0), the left-hand side has just one term, namely 2, "

What I dont understand is how was the left side of the equation narrowed downed to only 1 term which was 2.
What is the formula for the n-th term? What is the value of n when you start? So how many of the possible terms do you need?

Look at the examples in your book. They all have a general formula like the one they gave you here and they all do a base case where they only do one or a few terms. This is just like those.

SingingMommy
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Re: Mathematical Induction Question

Postby SingingMommy » Tue Nov 13, 2012 7:33 am

P(n) is the nth summation in the series. n = 0 would be P(0).

Each term is denoted as 2*(-7)^n. The first term is y(0) = 2*(-7)^0 = 2. Second term is y(1) = 2*(-7)^1 = -14. Third term is y(2) = 2*(-7)^2 = 98.
Since there is just one term so far, the sequence (adding up of the terms from n= 0, 1, 2, 3, ... , n) is just y(0).
P(0) = 2.

The first parts of the sequence is as follows:
P(0) = 2*(-7)^0 = 2 ; for n = 0.
P(1) = 2*(-7)^0 + 2*(-7)^1 = 2 + -14 = -12; This includes the terms for n = 0 and n = 1. It is summed up to n = 1.
P(2) = 2*(-7)^0 + 2*(-7)^1 + 2*(-7)^2 = 2 + -14 + 98 = 86; This includes the terms for n = 0, n = 1, and n = 2. it is summed up to n = 2.


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