What is the sum of all natural numbers $n$ such that the product of the digits of $n$ (in base $10$ ) is equal to $n^2-10 n-36 ?$
$12$
$13$
$124$
$2612$
Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$, with $\alpha>\beta$. For all positive integers $n$, define
$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$
$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$
Then which of the following options is/are correct?
$(1)$ $a_1+a_2+a_3+\ldots . .+a_n=a_{n+2}-1$ for all $n \geq 1$
$(2)$ $\sum_{n=1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$
$(3)$ $\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\frac{8}{89}$
$(4)$ $b=\alpha^n+\beta^n$ for all $n>1$
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