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Discrete Math: Induction |
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| Feb2-12, 09:43 PM | #1 |
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Discrete Math: Induction
Hi guys,
Long time lurker of this forum, but first time poster. Discrete Math is going to be the end of me; I'm just not understanding how to solve problems and write the proofs. Any help would be greatly appreciated. Thanks in advance. The Problem: Let nεZ≥1. Show that 1^3+3^3+5^3+.....+(2n-1)^3=n^2(2n^2-1). Proof: I'm using induction as this seems like a prime candidate. The claim was obvious, so I found myself at this next step: (2n-1)^3+(2(n+1)-1)^3=n^2(2n^2-1)+(2(n+1)-1)^3 From here, I just don't see how to algebraically manipulate one side to look like the other. I've been banging my head against the wall on this problem (and others) for hours now. Help! Thanks again! |
| Feb2-12, 09:43 PM | #2 |
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I decided to sanity check a few numbers and I think I found a problem, but I want to make sure that I am reading this right:
For 1, everything works: [2(1)-1)]^3=(1)^2[2(1)^2-1] 1^3=1^2*1^2 1=1 But I tried 7: [2(7)-1)]^3=(7)^2[2(7)^2-1] (13)^3=49(97) 2197=4753 |
| Feb2-12, 09:58 PM | #3 |
Recognitions:
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1^3+3^3+5^3+.....+(2n-1)^3=n^2(2n^2-1) that 1^3+3^3+5^3+.....+(2n-1)^3+(2(n+1)-1)^3=(n+1)^2(2(n+1)^2-1) right? The identity you want to prove is the difference between those two equations. |
| Feb2-12, 10:16 PM | #4 |
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Discrete Math: InductionThanks!!!!!! |
| Feb2-12, 10:22 PM | #5 |
Recognitions:
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| Feb3-12, 03:27 PM | #6 |
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Proof by Induction:
Show Base case is true. (n=1) Assume nth case is true. Want to show: (n+1)th case is true. Find relationship between n+1 case and n case and then use inequalities and equalities to prove it. |
| Feb3-12, 03:45 PM | #7 |
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For n= 7, you want [itex]1^3+ 3^3+ 5^3+ 7^3+ 9^3+ 11^3+ 13^3= 1+ 27+ 125+ 343+ 729+ 1331+ 2197= 4753[/itex] |
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