This solution isn't as elegant as could be... Anyway.
We have a recurrence relation
with initial condition
.
This recurrence is inhomogeneous (because of the
), but we can try homogenising it with a substitution
, where
is a constant.
We want the sequence
to satisfy
.
-(a_{n-1}+k)<br />\\ a_{n+1}&=14a_{n}-a_{n-1}+12k \Rightarrow 12k=-4 \Rightarrow k=-\frac{1}{3}<br />\end{alignedat})
Now we have
.
Solving our recurrence relation
with initial condition
, we obtain
(7+4\sqrt{3})^n+\frac{1}{6}(2+\sqrt{3})(7-4\sqrt{3})^n)
using characteristic equations and what-not (which I believe is quite standard in Olympiad Mathematics?).
Hence
(7+4\sqrt{3})^n+\frac{1}{6}(2+\sqrt{3})(7-4\sqrt{3})^n+\frac{1}{3})
, which is consistent with Alvin's post above.
(Alvin has
^n+\sqrt{3} \left(7-4 \sqrt{3}\right)^n+2 \left(7+4 \sqrt{3}\right)^n-\sqrt{3} \left(7+4 \sqrt{3}\right)^n\right))
.)
Now, from your progress on the problem, you observed that the sequence
seems to satisfy
with
.
If we define
to actually be the recurrence satisfying
with
, then solving the recurrence relation gives
(2+\sqrt{3})^n+\frac{1}{6}(3+\sqrt{3})(2-\sqrt{3})^n)
.
Now the only thing left to do is to cross our fingers and hope we can prove that
.
(2+\sqrt{3})^n+(3+\sqrt{3})(2-\sqrt{3})^n<br />\\ 36b_n^2&=(3-\sqrt{3})^{2}(2+\sqrt{3})^{2n}+(3+\sqrt{3})^{2}(2-\sqrt{3})^{2n}+2(3-\sqrt{3})(2+\sqrt{3})^n(3+\sqrt{3})(2-\sqrt{3})^n<br />\\ &= (12-6\sqrt{3})(2+\sqrt{3})^{2n}+(12+6\sqrt{3})(2-\sqrt{3})^{2n}+2(3-\sqrt{3})(3+\sqrt{3})((2-\sqrt{3})(2+\sqrt{3}))^n<br />\\ &= (12-6\sqrt{3})(2+\sqrt{3})^{2n}+(12+6\sqrt{3})(2-\sqrt{3})^{2n}+2\cdot6\cdot1^n<br />\\ &= (12-6\sqrt{3})(2+\sqrt{3})^{2n}+(12+6\sqrt{3})(2-\sqrt{3})^{2n}+12<br />\\ \Rightarrow b_n^2&=\frac{1}{36}(12-6\sqrt{3})(2+\sqrt{3})^{2n}+\frac{1}{36}(12+6\sqrt{3})(2-\sqrt{3})^{2n}+\frac{1}{3}<br />\\ &= \frac{1}{6}(2-\sqrt{3})(2+\sqrt{3})^{2n}+\frac{1}{6}(2+\sqrt{3})(2-\sqrt{3})^{2n}+\frac{1}{3}<br />\end{alignedat})
And finally,
^{2n}=((2\pm\sqrt{3})^2)^n=(7\pm4\sqrt{3})^n)
, so, indeed
.
The only other thing to prove is that
is actually an integer, which is easy, given that we defined it as the sequence
with
.
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