Let α and β be the roots of x ^ 2 - x - 1 = 0, with α > β . For all...

Given, $\alpha and$\beta$are roots of${x}^{2} - x - 1 = 0$</p><p>${a}_{k + 2} - {a}_{k} = \dfrac{\left({\alpha}^{k + 2} - {\beta}^{k + 2}\right) - \left({\alpha}^{k} - {\beta}^{k}\right)}{\alpha - \beta} = \dfrac{\left({\alpha}^{k + 2} - {\alpha}^{k}\right) - \left({\beta}^{k + 2} - {\beta}^{k}\right)}{\alpha - \beta}$</p><p>$= \dfrac{{\alpha}^{k}\left({\alpha}^{2} - 1\right) - {\beta}^{k}\left({\beta}^{2} - 1\right)}{\alpha - \beta} = {a}_{k + 1}$</p><p>$⇒ {a}_{k + 2} - {a}_{k} = {a}_{k + 1} ⇒ {a}_{k + 2} - {a}_{k + 1} = {a}_{k}$</p><p>$\mathop{∑ {a}_{k}}_{k = 1}^{n} = {a}_{k + 2} - {a}_{2} = {a}_{k + 2} - \dfrac{{\alpha}^{2} - {\beta}^{2}}{\alpha - \beta} = {a}_{k + 2} - \left(\alpha + \beta\right)$</p><p>$= {a}_{k + 2} - 1$</p><p>$⇒ {a}_{1} + {a}_{2} + {a}_{3} + . . . . . + {a}_{n} = {a}_{n + 2} - 1$<br />Option$( B )$<br />$\mathop{∑}_{n = 1}^{\infty}\dfrac{{a}_{n}}{{10}^{n}} = \mathop{∑}_{n = 1}^{\infty}\dfrac{{\alpha}^{n} - {\beta}^{n}}{( \alpha - \beta ) {10}^{n}}$<br />$= \dfrac{1}{\alpha - \beta}\mathop{∑}_{n = 1}^{\infty}\left({\left(\dfrac{\alpha}{10}\right)}^{n} - {\left(\dfrac{\beta}{10}\right)}^{n}\right)$<br />$= \dfrac{1}{\alpha - \beta}\left[\dfrac{\dfrac{\alpha}{10}}{1 - \dfrac{\alpha}{10}} - \dfrac{\dfrac{\beta}{10}}{1 - \dfrac{\beta}{10}}\right]$<br />$= \dfrac{1}{\alpha - \beta}\left[\dfrac{\alpha}{10 - \alpha} - \dfrac{\beta}{10 - \beta}\right]$<br />$= \dfrac{1}{\alpha - \beta}\left[\dfrac{10\alpha - 10\beta}{\left(10 - \alpha\right)\left(10 - \beta\right)}\right]$<br />$= \dfrac{1}{\alpha - \beta}\left[\dfrac{10\left(\alpha - \beta\right)}{100 - 10\left(\alpha + \beta\right) + \alpha\beta}\right]$<br />$= \dfrac{10}{100 - 10 - 1} = \dfrac{10}{89}$<br />Option$( C ) :$<br />$\mathop{∑}_{n = 1}^{\infty}\dfrac{{b}_{n}}{{10}^{n}} = \mathop{∑}_{n = 1}^{\infty}\dfrac{{\alpha}^{n} + {\beta}^{n}}{{10}^{n}} = \mathop{∑}_{n = 1}^{\infty}{\left(\dfrac{\alpha}{10}\right)}^{n} + {\left(\dfrac{\beta}{10}\right)}^{n}$<br />$= \dfrac{\dfrac{\alpha}{10}}{1 - \dfrac{\alpha}{10}} + \dfrac{\dfrac{\beta}{10}}{1 - \dfrac{\beta}{10}}$<br />apply sum of infinite G.P. with \left(a = \dfrac{\alpha}{10}andr = \dfrac{\alpha}{10}\right)$
$= \dfrac{10\left(\alpha + \beta\right) - 2\alpha\beta}{\left(10 - \alpha\right)\left(10 - \beta\right)}$
$= \dfrac{10 ( 1 ) - 2 ( - 1 )}{100 + \alpha\beta - 10 ( \alpha + \beta )} = \dfrac{12}{100 - 1 - 10} = \dfrac{12}{89}$
Option $( D ) :$
As given ${b}_{n} = {a}_{n - 1} + {a}_{n + 1}$
${b}_{n} = \dfrac{{\alpha}^{n - 1} - {\beta}^{n - 1}}{\alpha - \beta} + \dfrac{{\alpha}^{n + 1} - {\beta}^{n + 1}}{\alpha - \beta}$
$∵ \alpha - \beta = 1$
${b}_{n} = \left({\alpha}^{n - 1} - {\beta}^{n - 1}\right) + \left({\alpha}^{n + 1} - {\beta}^{n + 1}\right)$
Now, as $\alpha\beta = 1$
$∴ {\alpha}^{n}\beta = - {\alpha}^{n - 1}$
$∴ \alpha{\beta}^{n} = - {\beta}^{n - 1}$
${b}_{n} = {\alpha}^{n}\left(\alpha - \beta\right) + {\beta}^{n}\left(\alpha - \beta\right)$
${b}_{n} = {\alpha}^{n} + {\beta}^{n}$