## Mathematical Induction Question

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
ScripterKitty
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### Mathematical Induction Question

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

ScripterKitty wrote: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?

ScripterKitty wrote: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

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

ScripterKitty wrote: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"??

ScripterKitty wrote:"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

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.