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Cubic Sequence? (Posted on 2014-04-16) Difficulty: 3 of 5
107811/3, 110778111/3, 111077781111/3, 111107777811111/3, .....

Are all the terms in the infinite sequence given above perfect cubes?

If so, prove it. If not, provide a counterexample.

No Solution Yet Submitted by K Sengupta    
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Solution Easy to answer, proof not so much. (solution) | Comment 1 of 2
This is easy to answer, tedious to prove.
The answer is yes.  They are
33, 333, 3333, 33333, .....

If we add a first term to the beginning it would be
81/3 which is 3.

Then term n can be written as
[(10^(3n)-1)/9 - (10^(2n+1)-1)/9 + 7*(10^(2n)-1)/9 - 7*(10^(n)-1)/9 + (10^(n+1) - 1)/9]/3
=[10^(3n) - 10^(2n+1) + 7*10^(2n) - 7*10^(n) + 10^(n+1) - 1]/27

The sequence 3, 33, 333 ... can be written as
(10^(n)-1)/3
so the cubes are
[(10^(n)-1)/3] = [10^(3n)-3*10^(2n)+3*10^(n)-1]/27

these lead to the equation the must be verified
[10^(3n) - 10^(2n+1) + 7*10^(2n) - 7*10^(n) + 10^(n+1) - 1] = [10^(3n) - 3*10^(2n) + 3*10^(n) - 1]

to see this is true, note: 10^(n+1)-7*10^(n) = 3*10^(n)
and: - 10^(2n+1) + 7*10^(2n) = -3*10^(2n)
  Posted by Jer on 2014-04-17 10:55:34
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