super simple math equation I forgot

Factual Questions
1

Sorry, been a long time since high school and I can’t remember this simple equation.

I need to double the number one 64 times. In other words, 1x2 = 4. x2 = 8. x2 = 16. If I do that 64 times, what’s the total? And what’s the equation to figure it out?

2

Um, 1 x 2 = 2.

What you’re looking for is 2[sup]63[/sup]. 2[sup]0[/sup] = 1. 2[sup]1[/sup] = 2. 2[sup]2[/sup] = 4. 2[sup]63[/sup] is the 64th doubling.

9.22337204 × 10[sup]18[/sup]

3

2^64 … and 1x2 should be “2” in your example.

4

2[sup]64[/sup] would be the 64th doubling of 1. The first doubling is 1x2 =2, which is 2[sup]1[/sup].

5

Isn’t it 2[sup]64[/sup] = 18446744073709551616? First doubling is 2, second doubling is 4, so the nth doubling should be 2[sup]n[/sup].

6

Give him a break, it’s been even longer since elementary school.

If by “total” you mean all the numbers added together, as in the old story about a man who wanted one grain of rice for the first square on a chessboard, two for the second, four for the third, etc. the total of 2[sup]0[/sup] + 2[sup]1[/sup] + 2[sup]2[/sup]… + 2[sup]63[/sup] = 2[sup]64[/sup] - 1

7

Is that equation at the end correct? It seem somewhat counter-intuitive.

8

LOL oops. Does it help if I blame a screaming baby who refuses to go to sleep in the next room? No? Oh well.

Exactly what I am looking for!

9

Yes it is.

For example, try it on four squares:

2^0+2^1+2^2+2^3 = 1+2+4+8 = 15, which is 2^4-1.

10

Well, the formula must be right then! :rolleyes:

Maybe an actual proof will help the doubters?

  1. 2^0 = 1 = 2-1 = 2^1-1

  2. Suppose that for some n, 2^0 + 2^1 + … + 2^n = 2^(n+1) - 1
    Then 2^0 + 2^1 + … + 2^n + 2^(n+1) = 2^(n+1) - 1 + 2^(n+1)
    = 2*2(n+1)-1
    =2(n+2)-1

  3. Therefore, by induction, 2^0 + 2^1 + … + 2^n = 2^(n+1)-1 for all n.

11

Picture it in binary. You have a string of 64 consecutive 1s from 2[sup]0[/sup] all the way to 2[sup]63[/sup]. If you then added 1 to this you would flip every single bit as you carried 1 all the way from the right-hand end to the left, and a new 1 would appear at the left: 2[sup]64[/sup].

12

Did I ever claim I had a proof? :rolleyes:

Personally, I thought of the answer in Malacandra’s binary way, but I didn’t think that constitutes a proof, either, so I didn’t bother typing it.

13

Without induction:


x  = 2^0 + 2^1 + ... + 2^(n-1) + 2^n

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

2x = x - 1                     + 2^(n+1)

 x =   - 1 + 2^(n+1)

2^0 + 2^1 + ... + 2^(n-1) + 2^n = 2^(n+1) - 1