Numerator...Denominator...Terminator!

First of all, this is not an original problem, it is taken from some test and then edited a little bit.

After looking at the sequence, the only sequence which can(I guess) come in mind -

$\displaystyle D({T}_{n})= D({T}_{n-1}) + N({T}_{n-1})$

$\displaystyle N({T}_{n}) = D({T}_{n}) + D({T}_{n-1})$

But still, if there is any more series coming up, then please let me know I will provide this thing in the question only.

Let's just simply replace $D({T}_{n})$ by ${q}_{n}$ and $N({T}_{n})$ by ${p}_{n}$ .

$\displaystyle {q}_{n} = {q}_{n-1} + {p}_{n-1}$

$\displaystyle {p}_{n} = 2{q}_{n-1} + {p}_{n-1}$

Now, $\displaystyle {q}_{n} = 3{q}_{n-2} + 2{p}_{n-2}$

$\displaystyle {q}_{n} = 7{q}_{n-3} + 5{p}_{n-3}$

and so on, so we can write - $\displaystyle {q}_{n} = {a}_{k}{q}_{n-k} + {b}_{k}{p}_{n-k}$

As we can easily see, there is a recurrence going on in $\displaystyle {a}_{i}$ s and which is that -

$\displaystyle {a}_{k} = 2{a}_{k-1} + {a}_{k-2}$ with $\displaystyle {a}_{1} = 1$ , $\displaystyle {a}_{0} = 1$

And hence, the equation we will get is $\displaystyle {x}^{2} = 2x +1$

As we will bash a little bit more, we get $\displaystyle {a}_{k} = \frac{1}{2}({(1+\sqrt{2})}^{k} + {(1-\sqrt{2})}^{k})$

Also, $\displaystyle {b}_{k} = {b}_{k-1} + {a}_{k-1}$

$\displaystyle = \frac{1}{2}({(1+\sqrt{2})}^{k-1} + {(1-\sqrt{2})}^{k-1}) + {b}_{k-1}$

$\displaystyle = \frac{1}{2}({(1+\sqrt{2})}^{k-1} + {(1-\sqrt{2})}^{k-1}) + \frac{1}{2}({(1+\sqrt{2})}^{k-2} + {(1-\sqrt{2})}^{k-2}) + {b}_{k-2}$

And so on till $\displaystyle {b}_{0} = 0$ ,

$\displaystyle = \sum_{r=1}^{k}{\frac{1}{2}({(1+\sqrt{2})}^{k-r} + {(1-\sqrt{2})}^{k-r})}$

$\displaystyle = \frac{{(1+\sqrt{2})}^{k-1} - {(1-\sqrt{2})}^{k-1}}{2\sqrt{2}}$

Now, just plugging values and a little bashing, we get

$\displaystyle {q}_{n} = \frac{(2\sqrt{2}+1)({(1+\sqrt{2})}^{n-1}) + (2\sqrt{2}-1)({(1-\sqrt{2})}^{n-1})}{2\sqrt{2}}$ [Considering values of ${q}_{1}$ and ${p}_{1}$ ]

Well, let's move to ${p}_{n}$ which is defined as -

$\displaystyle {p}_{n} = 2{q}_{n-1} + {p}_{n-1}$

$\displaystyle {p}_{n} = \frac{(2\sqrt{2}+1)({(1+\sqrt{2})}^{n-1}) + (2\sqrt{2}-1)({(1-\sqrt{2})}^{n-1})}{\sqrt{2}} + {p}_{n-1}$

And as we know that $\displaystyle {p}_{1} = 1$ , we can bash a little bit again

$\displaystyle {p}_{n} = \frac{2 + (2\sqrt{2}+1)({(1+\sqrt{2})}^{n-1}-1) + (2\sqrt{2}-1)({(1-\sqrt{2})}^{n-1}-1)}{2}$

We have got equations of both - numerator as well as denominator and now comes the terminator which is just plugging in values.

$\displaystyle {p}_{29} = 99933150305$

$\displaystyle {p}_{17} = 2549185$

$\displaystyle {q}_{29} = 70663408246$

$\displaystyle {q}_{17} = 1802546$

$\displaystyle {lim}_{n->\infty}({T}_{n}) = \sqrt{2}$

Doing the final thing we get the answer as -

$\displaystyle \boxed{45}$