Question

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 $b_n={\alpha }_{n-1}+{\alpha }_{n+1},n\ge 2.$
Then which of the following options is/are correct?

• A. $a_1+a_2+a_3+-----+a_n=a_{n+2}-1$ for all $n\ge 1$
• B. $\sum _{n=1}^{\infty }\frac{a_n}{10^n}=\frac{10}{89}$
• C. $b_n={\alpha }^n+{\beta }^n$ for all $n\ge 1$
• D. $\sum _{n=1}^{\infty }\frac{b_n}{10^n}=\frac{8}{89}$

Answer 1

Answer: (A), (B), (C)

$\sum _{n=1}^{\infty }$ $\frac{a_n}{10^n}=$ $\sum _{n=1}^{\infty }$$\frac{{\alpha }^n-{\beta }^n}{\left(\alpha -\beta \right)10^n}$
$=\frac{1}{\alpha -\beta }\left[\sum _{n=1}^{\infty }{\left(\frac{\alpha }{10}\right)}^n-\sum _{n=1}^{\infty }{\left(\frac{\beta }{10}\right)}^n\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{\frac{\alpha }{10}}{1-\frac{\alpha }{10}}-\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{\alpha }{10-\alpha }-\frac{\beta }{10-\beta }\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{10\alpha -\alpha \beta -10\beta +\alpha \beta }{100-10\beta -10\alpha +\alpha \beta }\right]$
$=\frac{1}{\alpha -\beta }\left[\frac{10\left(\alpha -\beta \right)}{100-10-1}\right]=\frac{10}{89}$
$\left(B\right)a_{n+1}=\frac{{\alpha }^{n+1}-{\beta }^{n+1}}{\alpha -\beta }={\alpha }^n+{\alpha }^{n-1}\beta +{\alpha }^{n-2}{\beta }^2+...+\alpha .{\beta }^{n-1}+{\beta }^n$
$\left\{\because \alpha \beta =-1\right\}$
$a_{n+1}={\alpha }^n-\left({\alpha }^{n-2}+{\alpha }^{n-3}\beta +...+{\beta }^{n-2}\right)+{\beta }^n$
$\Rightarrow a_{n+1}={\alpha }^n+{\beta }^n-a_{n-1}$
$\Rightarrow a_{n-1}+{\alpha }_{n+1}={\alpha }^n+{\beta }^n$
$\Rightarrow b_n={\alpha }^n+{\beta }^n$
$\left(C\right)\because {\alpha }^2=\alpha +1and{\beta }^2=\beta +1$
${\alpha }^{n+2}={\alpha }^{n+1}+{\alpha }^{n}and{\beta }^{n+2}={\beta }^{n+1}+{\beta }^n$
${\alpha }^{n+2}-{\beta }^{n+2}=\left({\alpha }^{n+1}-{\beta }^{n+1}\right)+\left({\alpha }^n-{\beta }^n\right)$
$a_{n+2}=a_{n+1}+a_n...\left(i\right)$
Similarily $a_{n+1}=a_n+a_{n-1}...\left(ii\right)$
$a_n=a_{n-1}+a_{n-2}...\left(iii\right)$
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$a_3=a_2+a_1$
On adding $a_{n+2}=\left(a_1+a_2+a_3+...+a_n\right)+a_2\left(Here{a}_2=\alpha +\beta =1\right)$
$a_{n+2}-1=a_1+a_2+a_3+.......+a_n$
$\sum _{n=1}^{\infty }$$\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}$
$=\frac{\frac{\alpha }{10}}{1-\frac{\alpha }{10}}+\frac{\frac{\beta }{10}}{1-\frac{\beta }{10}}\because \left|\frac{\alpha }{10}\right|<1$
$\left|\frac{\beta }{10}\right|<1$
$=\frac{\alpha }{10-\alpha }+\frac{\beta }{10-\beta }$
$=\frac{10\alpha -\alpha \beta +10\beta -\alpha \beta }{\left(10-\alpha \right)\left(10-\beta \right)}$
$=\frac{10\left(1\right)-2\left(-1\right)}{100-10\left(\alpha +\beta \right)\alpha \beta }=\frac{12}{89}$