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September 24th, 2010, 09:58 AM  #1 
Newbie Joined: Sep 2010 Posts: 24 Thanks: 0  Help me prove something very easy! [Mathematical Induction]
Hey guys.. I'm new to MATH INDUCTION, so I wanted to ask you, to help me prove 2^n (greater than, or equal to) n^2 (where n is GREATER THAN or EQUAL to 4) I need DETAILED EXPLANATION pleaseeee I'm new, and I'm trying to do some problems, so I can get comfortable with it! Thank you! 
September 24th, 2010, 11:05 AM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5  Re: Help me prove something very easy! [Mathematical Induct
Let's call the statement P. You are trying to show that P(n) is true for all n>=4. Show that it is true for n = 4. Show that if it is true for a number, then it is true for the next number. 2^4=4^2, so P(4) is true. Assume P(n). Now the work is to show that P(n+1) is true. Have you done some simpler induction proofs, like 1+2+3+...+n = n(n+1)/2? 
September 24th, 2010, 12:18 PM  #3 
Newbie Joined: Sep 2010 Posts: 24 Thanks: 0  Re: Help me prove something very easy! [Mathematical Induct
Yeah yeah, I can do that without problem now.. I tried it many times! Now I fully understand that example! As for one I asked, I got to the n=k+1 step.. Now I can't seem to work them out, and get the form that I want them to be! I need some algebra there, some creativity.. that's why I'm trying to teach! If there are more simple problems like that, I'd like to know please! I'm searching for them all week long.. 
September 24th, 2010, 05:22 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Help me prove something very easy! [Mathematical Induct
2^n has to have 2^n added to it to get 2^(n + 1) while n^2 only has to have 2n + 1 added to get (n + 1)^2.

September 24th, 2010, 07:12 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Help me prove something very easy! [Mathematical Induct
2^n may be written as (1 + 1)^n = nCr(n,0) + nCr(n,1) + ... + nCr(n,n1) + nCr(n,n) When n ? 4 we have 2^n = 2n + 2 + [nCr(n,2) + ... + nCr(n,n2)] (at least one term in the brackets) It is easy to see that 2n + 2 + [nCr(n,2) + ... + nCr(n,n2)] ? 2n + 1 1 + [nCr(n,2) + ... + nCr(n,n2)] ? 0 
September 24th, 2010, 08:31 PM  #6  
Global Moderator Joined: Dec 2006 Posts: 21,027 Thanks: 2258  Quote:
Also, you should have been able to post your attempt to solve the problem before asking for help. There are numerous textbooks (some online) that explain how to use mathematical induction and give examples of its use.  

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