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

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

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Answer 1 · 2018-01-13T02:48:45

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#

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#