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Discussion: All Topics in PreCalculus for Mathematical Induction
Topic: Mathematical Induction


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Subject:   RE: Mathematical Induction
Author: Laura
Date: Nov 11 2003
Well, basically your hoework looks like this(from what I understand):
E(2^(i-1))=2^(n)-1.
Now, the E sigma sign means that i takes every value from 1 to n like this:
E(2^(i-1))=2^(1-1)+2^(2-1)+2^(3-1)+...+2^(n-1)=2^0+2^1+2^2+...+2^(n-1).
Now, from what you said, they have given you the answer, that is that the sum
equals 2^(n)-1, and you have to prove that it is corect. There are some ways
to do this, but by mathematical induction you do it like this:
There are 2 steps. The first one is to check if the relatioship is correct for
n=1 and n=2. That is check if:
E2^(i-1)=2^(n)-1 for n=1 and n=2.
that means: 2^0=2^1-1 and 2^0+2^1=2^2-1.(You have to do the math here)
The second step: you assume that the relationship is correct for n, and you
prove it for the value of n+1.
So:
You assume that E2^(i-1){i from 1 to n}=2^n-1
And you have to prove that
E2^(i-1){i from 1 to n+1}=2^(n+1)-1
You do thus by writing
E2^(i-1){i from 1 to n+1}=E2^(i-1){i from 1 to
n}+2^(n+1-1)=2^(n)-1+2^n=2^n+2^n-1=2^n(1+1)-1=2^(n)*2-1=2^(n+1)-1.
Thus, what you wanted to prove is proven.
For any given n, the relation works for n+1. (If it works for n=1, it will work
for n=2 the  for n=3 and so on).
Note that this is for this specific example. Any other one works the same(eg
E(i)=i(i+1)/2). The steps are exactly the same(check for n=1 and n=2, assume
it's corect for n and check for n+1), the only thing that changes is the
math.


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