Proof for result of sum of 3 elements of recursive sequence

Question

I have a recursive sequence: $$a_1=1\\a_2=1\\a_3=-1\\a_k=a_{k-1}\cdot a_{k-3} (for\,k>3)$$ So this sequence has cycle of 7: $1,1,-1,-1,-1,1,-1$
And I have to calc $a_{2013} + a_{2014} - a_{2015}\,\,$and the result is -3.
But I don't know how to prove it...
Thanks in advance for your help..

Answer 1

$$a_k=a_{k-1}\cdot a_{k-3}$$ $$=a_{k-2}\cdot a_{k-4}\cdot a_{k-3}$$ $$=a_{k-3}\cdot a_{k-5}\cdot a_{k-4}\cdot a_{k-3}$$ $$=a_{k-4}\cdot a_{k-5}$$ $$=a_{k-5}\cdot a_{k-7}\cdot a_{k-5}$$ $$=a_{k-7}, \forall k>7.$$ Thus, this sequence has cycle of $7$; the rest of the proof is easy.

Answer 2

$$a_4 = -1$$ $$a_5 = -1$$ $$a_6 = 1$$ $$a_7 = -1$$ $$a_8 = 1$$ $$a_9 = 1$$ $$a_{10} = -1$$ $$a_{11} = -1$$ $$a_{12} = -1$$ $$a_{13} = 1$$ $$a_{14} = -1$$

The sequence repeats in sets of 7.

So $a_{7k+4} = a_{2013}, a_{7k+5} = a_{2014}, a_{7k+6} = a_{2015}$.

Hence $a_{2013} = a_{2014} = -1$ and $a_{2015}=1$. $$a_{2013}+a_{2014}-a_{2015} = -1-1-1 = -3$$

Answer 3

As you correctly discovered, your sequence repeats after every seven steps. To prove it strictly, you can use recursion to prove that

$$a_{7k + 1} = 1\\ a_{7k + 2} = 1\\ a_{7k + 3} = -1\\ a_{7k + 4} = -1\\ a_{7k + 5} = -1\\ a_{7k + 6} = 1\\ a_{7k + 7} = -1\\ $$

for every value of $k\geq 0$. You can follow the standard steps in proving a statement using induction. You have to write a lot, but it isn't really all that hard.

A more informal proof (which I would still consider correct) would simply be:

Since $a_8=a_1, a_9=a_2$ and $a_{10} = a_3$, we can see that $a_{7+i}=a_i$ for every value of $i\geq 0$. From this, we can conclude that $a_{7k+i} = a_i$ for every $k\geq 0, i\geq 0$.

Once you prove that fact, you can calculate $$a_{2013} = a_{2009 + 4} = a_{7\cdot287 + 4} = -1\\ a_{2014} = a_{2009 + 5} = a_{7\cdot287 + 5} = -1\\ a_{2015} = a_{2009 + 6} = a_{7\cdot287 + 6} = 1\\ $$