Mathematical Induction

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>>328570
For the first expression,(2^n)-1/2^(n+1), It is just (1/2) * 2^(n-1)/(2^n). 2*(2^n) = 2^(n+1).
The second expression is just 1/2 in a form that you can use. Once again, 2*(2^n) = 2^(n+1) so 1/2 of 2^(n+1) is just 2^n.

Hope you understand this.

>>328574
Can you tell me what the rules used are?

>>328579
2*(2^n) = 2^(n+1)

>>328581
Where does the 2* in 2*(2^n) come from?

>>328582
From the denominator of 1/2.

So how is 2*(2^n) = 2^n+1?

>>328585
When you are multiplying two terms with the same base, e.g. 2 * 2^3 = 2^4, you add the exponents.

>>328585
https://en.wikipedia.org/wiki/Exponentiation#Identities_and_properties

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>>328588
How did I not notice? holy crap

So how does the second 1/2 become 2n/2^n+1

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>>328591
It's the same thing. Look at it again.

>>328591
2^n/2^n = 1 correct?
1/2 * 1 = 1/2
1/2 * (2^n/2^n) = 2^n/(2*2^n)
2^n/(2*2^n) = 2^n/(2^(n+1))

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>>328593
So the 1/2 was just ... ohhhh

How do I thank you for this?

>>328595
Learn to do math.