Let $\mathrm{\alpha }$ and $\mathrm{\beta }$ be the roots of ${\mathrm{x}}^2-\mathrm{x}-1=0,$ with $\mathrm{\alpha }>\mathrm{\beta }.$ For all positive integers $\mathrm{n},$ define
${\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{\alpha }}^{\mathrm{n}}-{\mathrm{\beta }}^{\mathrm{n}}}{\mathrm{\alpha }-\mathrm{\beta }},\mathrm{n}\ge 1$
${\mathrm{b}}_1=1$ and ${\mathrm{b}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}+1},\mathrm{n}\ge 2.$
Then which of the following options is/are correct?

Given, $\alpha$ and $\beta$ are roots of $x^2-x-1= 0$

$a_{k+2}-a_k=\frac{\left({\alpha }^{k+2}-{\beta }^{k+2}\right)-\left({\alpha }^k-{\beta }^k\right)}{\alpha -\beta }=\frac{\left({\alpha }^{k+2}-{\alpha }^k\right)-\left({\beta }^{k+2}-{\beta }^k\right)}{\alpha -\beta }$

$=\frac{{\alpha }^k\left({\alpha }^2-1\right)-{\beta }^k\left({\beta }^2-1\right)}{\alpha -\beta }=a_{k+1}$

$\Rightarrow a_{k+2}-a_k=a_{k+1 }\Rightarrow a_{k+2}-a_{k+1}=a_k$

$\underset{k=1}{\overset{n}{\sum a_k}}=a_{k+2}-a_{2 }=a_{k+2}-\frac{{\alpha }^2-{\beta }^2}{\alpha -\beta }=a_{k+2}-\left(\alpha +\beta \right)$

$=a_{k+2}-1$

$\Rightarrow {\mathrm{a}}_1+{\mathrm{a}}_2+{\mathrm{a}}_3+.....+{\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_{\mathrm{n}+2}-1$ 
Option $\left(\mathrm{B}\right)$
$\sum _{n=1}^{\infty }\frac{a_n}{{10}^n}=\sum _{n=1}^{\infty }\frac{{\alpha }^n-{\beta }^n}{(\alpha -\beta ){10}^n}$
$=\frac{1}{\alpha -\beta }\sum _{n=1}^{\infty }\left({\left(\frac{\alpha }{10}\right)}^n-{\left(\frac{\beta }{10}\right)}^n\right)$
$=\frac{1}{\alpha -\beta }\left[\frac{\frac{\alpha }{10}}{1-\frac{\alpha }{10}}-\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta }\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{10\alpha -10\beta }{\left(10-\alpha \right)\left(10-\beta \right)}\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{10\left(\alpha -\beta \right)}{100-10\left(\alpha +\beta \right)+\alpha \beta }\right]$
$=\frac{10}{100-10-1}=\frac{10}{89}$
Option $\left(\mathrm{C}\right):$
$\sum _{n=1}^{\infty }\frac{b_n}{{10}^n}=\sum _{n=1}^{\infty }\frac{{\alpha }^n+{\beta }^n}{{10}^n}=\sum _{n=1}^{\infty }{\left(\frac{\alpha }{10}\right)}^n+{\left(\frac{\beta }{10}\right)}^n$
$=\frac{{\displaystyle \frac{\alpha }{10}}}{1-\frac{\alpha }{10}}+\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}$
apply sum of infinite G.P. with $\left(a=\frac{\alpha }{10}\mathrm{a}\mathrm{n}\mathrm{d}r=\frac{\alpha }{10}\right)$
$=\frac{10\left(\alpha +\beta \right)-2\alpha \beta }{\left(10-\alpha \right)\left(10-\beta \right)}$
$=\frac{10\left(1\right)-2(-1)}{100+\alpha \beta -10(\alpha +\beta )}=\frac{12}{100-1-10}=\frac{12}{89}$
Option $\left(\mathrm{D}\right):$
As given $b_n=a_{n-1}+a_{n+1}$
$b_n=\frac{{\alpha }^{n-1}-{\beta }^{n-1}}{\alpha -\beta }+\frac{{\alpha }^{n+1}-{\beta }^{n+1}}{\alpha -\beta }$
$\because \alpha -\beta =1$
$b_n=\left({\alpha }^{n-1}-{\beta }^{n-1}\right)+\left({\alpha }^{n+1}-{\beta }^{n+1}\right)$
Now, as $\alpha \beta = 1$
$\therefore {\alpha }^n\beta =-{\alpha }^{n-1}$
$\therefore \alpha {\beta }^n=-{\beta }^{n-1}$
$b_n={\alpha }^n\left(\alpha -\beta \right)+{\beta }^n\left(\alpha -\beta \right)$
$b_n={\alpha }^n+{\beta }^n$