Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$ with $\alpha >\beta$ . For all positive integers n, define $a_n=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },n\ge 1$, $b_1=1$ and $b_n=a_{n-1}+a_{n+1},n\ge 2$ Then which of the following options is/are correct?

$\begin{array}{l}\text{Given }\alpha ,\beta \text{\hspace{0.17em} are the roots of the equation }{x}^2-x-1=0\\ \text{hence, }\alpha =\frac{1+\sqrt{5}}{2}, \beta =\frac{1-\sqrt{5}}{2}\end{array}$

$\begin{array}{c}\text{Given }{a}_n=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },n\ge 1\\ Given,b_1=1\text{\hspace{0.17em}\hspace{0.17em}and }\\ b_n=a_{n-1}+a_{n+1}\\ =\frac{{\alpha }^{n-1}-{\beta }^{n-1}}{\alpha -\beta }+\frac{{\alpha }^{n+1}-{\beta }^{n+1}}{\alpha -\beta }\\ =\frac{{\alpha }^{n-1}\left({\alpha }^2+1\right)-{\beta }^{n-1}\left({\beta }^2+1\right)}{\alpha -\beta }\\ =\frac{{\alpha }^{n-1}\left(\alpha +2\right)-{\beta }^{n-1}\left(\beta +2\right)}{\alpha -\beta }\\ =\frac{{\alpha }^{n-1}\left(\frac{5+\sqrt{5}}{2}\right)-{\beta }^{n-1}\left(\frac{5-\sqrt{5}}{2}\right)}{\sqrt{5}}\\ ={\alpha }^{n-1}\left(\alpha \right)-{\beta }^{n-1}\left(-\beta \right)={\alpha }^n+{\beta }^n\\ \text{and }\sum _{n=1}^{\infty }\frac{b_n}{{10}^n}=\sum _{n=1}^{\infty }\frac{{\alpha }^n+{\beta }^n}{{10}^n}=\sum _{n=1}^{\infty }{\left(\frac{\alpha }{10}\right)}^n+\sum _{n=1}^{\infty }{\left(\frac{\beta }{10}\right)}^n\end{array}$

$\begin{array}{l}=\frac{\frac{\alpha }{10}}{1-\frac{\alpha }{10}}+\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}=\frac{\alpha }{10-\alpha }+\frac{\beta }{10-\beta }=\frac{10(\alpha +\beta )-2\alpha \beta }{100-10(\alpha +\beta )+\alpha \beta }=\frac{10\left(1\right)-2\left(-1\right)}{100-10+(-1)}=\frac{12}{89}\\ \sum _{n=1}^{\infty }\frac{a_n}{{10}^n}=\sum _{n=1}^{\infty }\frac{{\alpha }^n-{\beta }^n}{(\alpha -\beta ){10}^n}=\frac{1}{\sqrt{5}}\left(\sum _{n=1}^{\infty }{\left(\frac{\alpha }{10}\right)}^n-\sum _{n=1}^{\infty }{\left(\frac{\beta }{10}\right)}^n\right)=\frac{1}{\sqrt{5}}\left(\frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta }\right)\\ =\frac{1}{\sqrt{5}}\left(\frac{10(\alpha -\beta )}{100-10(\alpha +\beta )+\alpha \beta }\right)=\frac{1}{\sqrt{5}}\left(\frac{10\times \sqrt{5}}{89}\right)=\frac{10}{89} \text{ Also }{a}_1+a_2+a_3+\dots \dots .+a_n\\ =\frac{1}{\alpha -\beta }\left[(\alpha -\beta )+\left({\alpha }^2-{\beta }^2\right)+\left({\alpha }^3-\beta \right)+\dots \dots .+\left({\alpha }^n-{\beta }^n\right)\right]\\ =\frac{1}{\alpha -\beta }\left[\left(\alpha +{\alpha }^2+\dots ..+{\alpha }^n\right)-\left(\beta +{\beta }^2+\dots \dots {\beta }^n\right)\right]\\ =\frac{1}{\alpha -\beta }\left[\left(1+\alpha +{\alpha }^2\dots \dots +{\alpha }^n\right)-\left(1+\beta +{\beta }^2\dots \dots .{\beta }^n\right)\right]\\ =\frac{1}{\alpha -\beta }\left[\frac{{\alpha }^{n+1}-1}{\alpha -1}-\frac{{\beta }^{n+1}-1}{\beta -1}\right]=\frac{{\alpha }^{n+2}-{\beta }^{n+2}}{\alpha -\beta }-1=a_{n+2}-1\end{array}$