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#1 2009-10-07 22:35:22

morgandebbie
Member
Registered: 2009-10-07
Posts: 1

induction

Can somebody help me with proving:

thanx

Last edited by morgandebbie (2009-10-07 23:11:11)

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#2 2009-10-08 07:18:14

Induction
Guest

Re: induction

first  based on rules:
1)n=1 (a^1-b^1)a-b)\sum_{i=0}^{}(a^ib^{-i})

#3 2009-10-08 07:21:17

G_Einstein
Member
Registered: 2008-08-30
Posts: 124

Re: induction

first baesd on rules:
1)n=1


2)n=k

3)n=k+1

not sure if that a "proof"

Last edited by G_Einstein (2009-10-08 07:33:09)


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#4 2009-10-27 08:38:01

ahed
Member
Registered: 2009-10-27
Posts: 2

Re: induction

Suppose  f : Z+  Z+ is defined so that f(1) = 2, and f(n+1) = f(n) + 2(n + 1). Prove by induction that f(n) = n(n + 1).

whats the answer of this please

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#5 2009-10-27 12:37:54

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: induction

Hi ahed;

Suppose  f : Z+  Z+ is defined so that

I can't read this:

You can prove this inductively like this;

1) Prove for the base case:

f(1) = 2 = 1(1+1) is true.

This is the assertion: f(n) = n(n+1)

This is the inductive step:

f(n+1) = (n+1)((n+1) + 1)= (n+1)(n+2)

Take the inductive step recurrence:

f(n+1) = f(n) + 2(n+1)

Plug in the assertion by substituting for f(n) = n(n+1)

f(n+1) = n(n+1) + 2(n+1) = (n+1)(n+2) This equals the inductive step so the assertion is proved by induction.

Last edited by bobbym (2009-10-28 03:26:00)


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#6 2009-10-28 01:43:41

ahed
Member
Registered: 2009-10-27
Posts: 2

Re: induction

thanks

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