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View Full Version : Actuarial Review (Nov. 02) It's a Puzzlement!

MathGuy
11-21-2002, 01:36 PM
Anyone else working on the Puzzlement from the current AR? It involves placing ten 30-60-90 triangles, with shortest sides 1, 2, 2, 3, 3, 3, 4, 4, 4, and 4 inside another 30-60-90 triangle with shortest side 10. I've created a nifty little spreadsheet that lets me move the triangles around, along with rotation and flipping. Right now, I think that the visual "tangram" method is the best way to solve it. But we'll see...

Anyway, I went ahead and proved John Robertson's statement of the "nonobvious result" that:

1<sup>3</sup> + 2<sup>3</sup> + ... + n<sup>3</sup> = (1 + 2 + ... + n)<sup>2</sup>.

My proof is by induction and requires the fact that 1 + ... + n = .5(n<sup>2</sup> - n).

First, we note that:

1<sup>3</sup> = 1<sup>2</sup>
1<sup>3</sup> + 2<sup>3</sup> = (1 + 2)<sup>2</sup>

Now, we assume that this is true for all integers up to k-1, namely that:

1<sup>3</sup> + 2<sup>3</sup> + ... + (k-1)<sup>3</sup> = (1 + 2 + ... + k-1)<sup>2</sup>.

And we use this to show that it is true for k:

(1 + ... + k)<sup>2</sup>
= (1 + ... + k)(1 + ... + k)
= ([1 + ... + k-1] + k)([1 + ... + k-1] + k)
= [1 + ... + k-1]<sup>2</sup> + 2k[1 + ... + k-1] + k<sup>2</sup>
= 1<sup>3</sup> + 2<sup>3</sup> + ... + (k-1)<sup>3</sup> +2k(.5)(k<sup>2</sup> - k) + k<sup>2</sup>
= 1<sup>3</sup> + 2<sup>3</sup> + ... + (k-1)<sup>3</sup> + k<sup>3</sup> - k<sup>2</sup> + k<sup>2</sup>
= 1<sup>3</sup> + 2<sup>3</sup> + ... + (k-1)<sup>3</sup> + k<sup>3</sup>.

And we're all set.

Cho Da
11-21-2002, 02:20 PM
... the fact that 1<sup>2</sup> + ... + n<sup>2</sup> = .5(n<sup>2</sup> - n).
:-?

1¹ + … + n¹ = ½(n² + n )

thing
11-21-2002, 03:05 PM
From how he used it in the proof, that's what he meant to say...

MathGuy
11-21-2002, 03:08 PM
Yeah, sorry, got caught up in my notation. It's now fixed.

Cho Da
11-21-2002, 03:14 PM
Yeah, sorry, got caught up in my notation. It's now fixed.But have you fit the triangles together yet?

Spike
11-21-2002, 03:24 PM
To continue the quibbling, it's actually

1 + … + (n-1) = ½(n² - n )

MathGuy
11-21-2002, 03:33 PM
Actually, what I meant was 1 + ... + n = .5(n<sup>2</sup> + n).

MathGuy
11-21-2002, 03:42 PM
Anyway, I went ahead and proved John Robertson's statement of the "nonobvious result" that:

1<sup>3</sup> + 2<sup>3</sup> + ... + n<sup>3</sup> = (1 + 2 + ... + n)<sup>2</sup>.

My proof is by induction and requires the fact that 1 + ... + n = .5(n<sup>2</sup> + n).

First, we note that:

1<sup>3</sup> = 1<sup>2</sup>
1<sup>3</sup> + 2<sup>3</sup> = (1 + 2)<sup>2</sup>

Now, we assume that this is true for all integers up to k, namely that:

1<sup>3</sup> + 2<sup>3</sup> + ... + k<sup>3</sup> = (1 + 2 + ... + k)<sup>2</sup>.

And we use this to show that it is true for k+1:

(1 + ... + k+1)<sup>2</sup>
= (1 + ... + k+1)(1 + ... + k+1)
= ([1 + ... + k] + k+1)([1 + ... + k] + k+1)
= [1 + ... + k]<sup>2</sup> + 2(k+1)[1 + ... + k] + (k+1)<sup>2</sup>
= 1<sup>3</sup> + ... + k<sup>3</sup> +2(k+1)(.5)(k<sup>2</sup> + k) + (k+1)<sup>2</sup>
= 1<sup>3</sup> + ... + k<sup>3</sup> + (k+1)(k<sup>2</sup>+k) + (k+1)<sup>2</sup>
= 1<sup>3</sup> + ... + k<sup>3</sup> + (k+1)(k+1)k + (k+1)<sup>2</sup>
= 1<sup>3</sup> + ... + k<sup>3</sup> + (k+1)<sup>2</sup>k + (k+1)<sup>2</sup>
= 1<sup>3</sup> + ... + k<sup>3</sup> + (k+1)<sup>2</sup>(k+1)
= 1<sup>3</sup> + ... + k<sup>3</sup> + (k+1)<sup>3</sup>.