Question

How to find the sum of `1+3+3^2+...+3^(3^3)`?

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

Answer 1

Set the sum to a number `S`, multiply that sum by `3` and subtract the original sum from this new sum, then simplify to get

`S = (3^28 - 1)/2 = 11438396227480`

Explanation:

Let's set the sum to a number `S`:

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

That `3^(3^3)` in the end could be simply `3^27`:

`S = 1 + 3 + 3^2 + ... + 3^27`

Now let's "reverse" the sum:

`S = 3^27 + 3^26 + ... + 3^2 + 3 + 1`

Multiply it by `3`:

`3S = 3^28 + 3^27 + ... + 3^3 + 3^2 + 3`

And subtract the original sum from this:

`3S - S = (3^28 + 3^27 + 3^26 + ... + 3^2 + 3) - (3^27 + 3^26 + ... + 3^2 + 3 + 1)`

Almost all of the terms cancel out:

`2S = 3^28 - 1`

And we can simply divide by `2`:

`S = (3^28 - 1)/2`

As for what `3^28` is, we could use a calculator and get

`S = (22876792454961 - 1)/2 = (22876792454960)/2 = 11438396227480`

Therefore, our sum is equal to:

`S = 1 + 3 + 3^2 + ... + 3^(3^3) = (3^28 - 1)/2 = 11438396227480`