How can induction be used to prove a sum of cubes formula?

[hammonjj]

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!

 

[hammonjj]

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

 

[Dick]

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!

Welcome to the forums! You'll need to show how you decided that was the thing you need to prove, especially since it's not true. What you want to show is that given:

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.

 

[hammonjj]

Welcome to the forums! You'll need to show how you decided that was the thing you need to prove, especially since it's not true. What you want to show is that given:

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.

Thanks for the help so far. Apparently, I'm a little on the slow side from banging my head on the wall from this. Can you go an extra step or two because I still don't know how to solve this.

Thanks!

 

[Dick]

Thanks for the help so far. Apparently, I'm a little on the slow side from banging my head on the wall from this. Can you go an extra step or two because I still don't know how to solve this.

Thanks!

Do you agree with what I've said so far? The first line is nth case of what you are supposed to show and the second is the (n+1)th. You are supposed to show the first implies the second. That's induction. Subtract the first from the second. What do you get?

 

[Shayes]

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.

 

[HallsofIvy]

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

Yes, it is NOT true, in general, that $(2n-1)^3= n^2(2n^2-1)$ but that is NOT what you are asked to prove. The left side is a sum of cubes, not just a single cube.

For n= 7, you want
$1^3+ 3^3+ 5^3+ 7^3+ 9^3+ 11^3+ 13^3= 1+ 27+ 125+ 343+ 729+ 1331+ 2197= 4753$