Actuarial Review (Nov. 02) It's a Puzzlement!

[MathGuy]

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:

1³ + 2³ + ... + n³ = (1 + 2 + ... + n)².

My proof is by induction and requires the fact that 1 + ... + n = .5(n² - n).

First, we note that:

1³ = 1²
1³ + 2³ = (1 + 2)²

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

1³ + 2³ + ... + (k-1)³ = (1 + 2 + ... + k-1)².

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

(1 + ... + k)²
= (1 + ... + k)(1 + ... + k)
= ([1 + ... + k-1] + k)([1 + ... + k-1] + k)
= [1 + ... + k-1]² + 2k[1 + ... + k-1] + k²
= 1³ + 2³ + ... + (k-1)³ +2k(.5)(k² - k) + k²
= 1³ + 2³ + ... + (k-1)³ + k³ - k² + k²
= 1³ + 2³ + ... + (k-1)³ + k³.

And we're all set.

[Cho Da]

... the fact that 1² + ... + n² = .5(n² - n).
:-?

1¹ + … + n¹ = ½(n² + n )

[thing]

From how he used it in the proof, that's what he meant to say...

[MathGuy]

Yeah, sorry, got caught up in my notation. It's now fixed.

[Cho Da]

Yeah, sorry, got caught up in my notation. It's now fixed.But have you fit the triangles together yet?

[Spike]

To continue the quibbling, it's actually

1 + … + (n-1) = ½(n² - n )

[MathGuy]

Actually, what I meant was 1 + ... + n = .5(n² + n).

Please wait...

[MathGuy]

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

1³ + 2³ + ... + n³ = (1 + 2 + ... + n)².

My proof is by induction and requires the fact that 1 + ... + n = .5(n² + n).

First, we note that:

1³ = 1²
1³ + 2³ = (1 + 2)²

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

1³ + 2³ + ... + k³ = (1 + 2 + ... + k)².

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

(1 + ... + k+1)²
= (1 + ... + k+1)(1 + ... + k+1)
= ([1 + ... + k] + k+1)([1 + ... + k] + k+1)
= [1 + ... + k]² + 2(k+1)[1 + ... + k] + (k+1)²
= 1³ + ... + k³ +2(k+1)(.5)(k² + k) + (k+1)²
= 1³ + ... + k³ + (k+1)(k²+k) + (k+1)²
= 1³ + ... + k³ + (k+1)(k+1)k + (k+1)²
= 1³ + ... + k³ + (k+1)²k + (k+1)²
= 1³ + ... + k³ + (k+1)²(k+1)
= 1³ + ... + k³ + (k+1)³.