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**GGPaltrow** Having a finite sequence of numbers given, we create a new sequence by inputting in each step between every pair of two adjacent numbers a new number equal to their sum. We start with (1,1), in the second step we have (1,2,1), in the third (1,3,2,3,1) etc. For every $\displaystyle n\geq1$ calculate the sum of the cubes of the numbers being part of the sequence acquired in the nth step.

What we know is that in every step, for a sequence of lenght n we'll get n-1 new numbers being the sums of the adjacents so the next sequence will be 2n-1. The sum of the first is $\displaystyle 1^3+1^3=1$, then we have $\displaystyle 1^{3}+2^{3}+1^{3}=1+8=9$, then the_sum_so_far+$\displaystyle 2*3^{3}$. The useful property is that the sequence is symmetrical having some k pairs of numbers on both sides of the central 2 and always has an odd amount of numbers - only the first step is even. How does that lead to a solution, though? Could you please help?