Question

Let $\alpha;$ and $\beta$ be the roots of $x^2 — x — 1 = 0$, with $\alpha > \beta.$ For all positive integers n, define $a_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta}, n\ge1,$
$b_{1}=1$ and $b_{n}=\alpha_{n-1}+\alpha_{n+1}, n\ge2.$
Then which of the following options is/are correct?

• A. $a_1 + a_2 + a_3 +-----+ a_n = a_{n+2} - 1$ for all $n\ge1$
• B. $\displaystyle \sum_{n=1}^\infty \frac{a_n}{10^n} =\frac{10}{89}$
• C. $b_{n}=\alpha^{n}+\beta^{n}$ for all $n\ge1$
• D. $\displaystyle \sum_{n=1}^\infty \frac{b_n}{10^n} =\frac{8}{89}$

Answer 1

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

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