# Thread: Sequence cube sums

1. ## Sequence cube sums

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?

2. ## Re: Sequence cube sums

Originally Posted by 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?

Hello!

First use [TEX] tabs and not [tex].

...and $\displaystyle 1^3+1^3=1+1=2$

Hint:

Binomial coefficient - Wikipedia, the free encyclopedia

5. ## Re: Sequence cube sums

Thank you very much. Though this example seems similar to the Pascal's triangle shown in the binomial coefficient, it's not exactly that... The second link doesn't work for me. The third is interesting as the consecutive franel numbers are quite close to the sums of the cubes in this example! How to put this in use, though?

6. ## Re: Sequence cube sums

OK, after some playing with numbers, I determined the sum will be $\displaystyle 9*7^{n-2}+1$ for $\displaystyle n\geq2$ but have no idea on how to prove this... Could you please help?

7. ## Re: Sequence cube sums

I'm trying to prove this by induction but it doesn't seem to nail the trick. Could you?