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.
This is easy to answer, tedious to prove.
The answer is yes. They are
33³, 333³, 3333³, 33333³, .....
If we add a first term to the beginning it would be
81/3 which is 3³.
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]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 20140417 10:55:34 