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.

(In reply to Easy to answer, proof not so much. (solution) by Jer)

I agree.

Let x = 1/3(10^(n+1)-1) {33,333,3333,33333,...} clearly a whole number.       
Then x^3 = (1/3 (-1+10^(1+n)))^3 = 1/27 (10^(n+1)-1)^3, clearly still a whole number.       
And 3x^3 = 1/9 (10^(n+1)-1)^3 {107,811, 110,778,111, 111,077,781,111,...} still whole numbers,       
But also (essentially, transcribing and simplifying):  1/9*(10^n-1)*10^(2n+3)+7/9*(10^n-1)*10^(n+2)+8*10^(n+1)+10/9*(10^n-1)+1

= 1/9(10^(n+1)-1)^3 

of which we take 1/3 to obtain 1/27(10^(n+1)-1)^3 = (1/3(10^(n+1)-1))^3; the cube of the series {33,333,3333,33333,...}, as was to be shown.

 

Edited on April 18, 2014, 3:50 am

Posted by broll on 2014-04-17 15:47:39