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### Triangular Numbers in a Proof

```
Date: 04/08/97 at 19:17:43
From: Suzanne Flanagan
Subject: Triangular numbers in a proof

1^3 = 1^2
1^3 +2^3 = (1+2)^2
1^3 +2^3 +3^3 = (1+2+3)^2
1^3 +2^3 +3^3+ ...+n^3 = (1+2+3+...+n)^2.

I am trying to prove
1^3 = 1^2
1^3 +2^3 = (1+2)^2
1^3 +2^3 +3^3 = (1+2+3)^2
1^3 +2^3 +3^3+ ...+n^3 = (1+2+3+...+n)^2.

I have tried to find a proof by induction, but didn't get very far.
I also tried working with triangular numbers since the right side is
the triangular numbers, but I could not show that the left and right
sides were equal. I need some hints, or maybe the trick that is needed
to verify the general case. I also found some patterns which turned
out to lead to a dead end.

Please HELP!
Suzanne Flanagan
```

```
Date: 04/13/97 at 12:34:16
From: Doctor Steven
Subject: Re: Triangular numbers in a proof

The right hand side is equal to

[(n*(n+1))/2]^2

so you are trying to prove that the sums of the cubes of numbers is
equal to

(n^4 + 2n^3 + n^2)/4.

Induction should work fairly well for this proof.

Start with your base case of 1:

(1^4 + 2*1^3 + 1^2)/4 = 1^3 = 1.

Assume it's true for k:

(k^4 + 2k^3 + k^2)/4 = 1^3 + 2^3 + .... + k^3.

Look at k + 1:

((k+1)^4 + 2*(k+1)^3 + (k+1)^2)/4 =
(k^4 + 4k^3 + 6k^2 + 4k + 1 + 2k^3 + 6k^2 + 6k + 2 + k^2 + 2k + 1)/4=
(k^4 + 6k^3 + 13k^2 +  12k + 4)/4 =
(k^4 + 2k^3 + k^2)/4 + (4k^3 + 12k^2 + 12k + 4)/4 =
(1^3 + 2^3 + 3^3 + . . . + k^3) + (k^3 + 3k^2 + 3k + 1) =
(1^3 + 2^3 + 3^3 + . . . + k^3) + (k + 1)^3 =
(1^3 + 2^3 + 3^3 + . . . + k^3 + (k+1)^3.

We win!

Hope this helps.

-Doctor Steven,  The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/09/2001 at 23:14:56
From: Madison
Subject: Mathematical induction

Prove by mathematical induction:

1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2
```

```
Date: 10/10/2001 at 07:11:21
From: Doctor Floor
Subject: Re: Mathematical induction

Hi, Madison,

It is easy to check that this is correct for n = 1. Suppose that we
have proven it is correct that 1^3+2^3+3^3+...+n^3 = (1+2+3+...+n)^2,
which we will use as induction hypothesis. We will use the well-known
identity for triangular numbers

1+2+3+...+n = n(n+1)/2

See for instance, from the Dr. Math archives:

Formula for Triangular Numbers
http://mathforum.org/dr.math/problems/huff8.29.98.html

Then we can derive the following:

1^3+2^3+3^3+...+n^3+(n+1)^3
= (1+2+3+...+n)^2+(n+1)^3
= (1+2+3+...+n)^2 + n(n+1)^2 +(n+1)^2
= (1+2+3+...+n)^2 + n(n+1)*(n+1) + (n+1)^2
= (1+2+3+...+n)^2 + 2(1+2+3+...+n)*(n+1) + (n+1)^2
= (1+2+3+...+n + n+1)^2

And the proof by induction is complete.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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