A probability problem by Sourabh Jangid

From the recurrence relation $a_{n+2} = 2a_{n+1}+a_n$ , the characteristic equation is $r^2 - 2r -1 = 0$ $\implies r = \dfrac {2 \pm \sqrt{4+4}}2 = 1 \pm \sqrt 2$ .

Therefore, we have:

$\begin{aligned} a_n & = c_1(1+\sqrt 2)^n + c_2(1-\sqrt 2)^n \\ a_0 & = c_1 + c_2 = 1 \\ a_1 & = c_1(1+\sqrt 2) + c_2(1-\sqrt 2) \\ \implies 1 & = c_1 + c_2 + \sqrt 2(c_1-c_2) \\ 1 & = 1 + \sqrt 2(c_1-c_2) \\ \implies c_1 & = c_2 = \frac 12 \\ \implies a_n & = \frac 12 \left((1+\sqrt 2)^n + (1-\sqrt 2)^n \right) \end{aligned}$

Then, we have:

$\begin{aligned} \sum_{n=0}^\infty \frac {a_n}{5^{2n}} & = \frac 12 \sum_{n=0}^\infty \left[\left(\frac {1+\sqrt 2}{25}\right)^n + \left(\frac {1-\sqrt 2}{25}\right)^n \right] \\ & = \frac 12 \left[\frac 1{1-\frac {1+\sqrt 2}{25}} +\frac 1{1-\frac {1-\sqrt 2}{25}} \right] \\ & = \frac 12 \left[\frac {25}{24 - \sqrt 2} + \frac {25}{24 + \sqrt 2} \right] \\ & = \frac {25}2 \left[\frac {24 + \sqrt 2 + 24 - \sqrt 2}{(24 - \sqrt 2)(24 + \sqrt 2)} \right] \\ & = \frac {25}2 \cdot \frac {48}{24^2-2} \\ & = \frac {600}{574} = \frac {599+1}{574} \end{aligned}$

$\implies \alpha = \boxed{1}$

Applying Z transform and after some work, one realize that:

$\large a(n) = \frac12[(1+\sqrt{2})^n + (1-\sqrt{2})^n]$

So

$\large \sum_{n=0}^{\infty} \frac{a_n} {5^{2n}}$

Is:

$\large \frac12 \Big [\sum_{n=0}^{\infty} (\frac{1+\sqrt{2}} {25})^n + \sum_{n=0}^{\infty} (\frac{1-\sqrt{2}} {25})^n \Big]$

$\large = \frac12 \Big[\frac{1}{1 - \frac{1 + \sqrt{2}}{25}} + \frac{1}{1 - \frac{1 - \sqrt{2}}{25}} \Big]$

$\large = \frac12 \Big( \frac{25}{24 - \sqrt{2}} + \frac{25}{24 - \sqrt{2}} \Big)$

$\large =\frac{600}{574}$

So:

$\color{#3D99F6} \alpha = 1$