Given quadratic equation

$${x^2} - x - 1 = 0$$ having roots $$\alpha $$ and $$\beta $$, ($$\alpha $$ > $$\beta $$)

So, $$\alpha = {{1 + \sqrt 5 } \over 2}$$ and $$\beta = {{1 - \sqrt 5 } \over 2}$$

and $$\alpha + \beta = 1$$, $$\alpha $$$$\beta $$ = $$ - 1$$

$$ \because $$ $${a_n} = {{{\alpha ^n} - {\beta ^n}} \over {\alpha - \beta }},\,n \ge 1$$

So, $${a_{n + 1}} = {{{\alpha ^{n + 1}} - {\beta ^{n + 1}}} \over {\alpha - \beta }}$$

$${\alpha ^n} + {\alpha ^{n - 1}}\beta + {\alpha ^{n - 2}}{\beta ^2} + ... + \alpha {\beta ^{n - 1}} + {\beta ^n}$$

$${\alpha ^n} - {\alpha ^{n - 2}} - {\alpha ^{n - 3}}\beta - ... - {\beta ^{n - 2}} + {\beta ^n}$$

[as $$\alpha $$$$\beta $$ = $$ - $$1]

= $${\alpha ^n} + {\beta ^n} - ({\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}})$$

= $${\alpha ^n} + {\beta ^n} - {a_{n - 1}}$$

$$\left[ {as\,{a_{n - 1}} = {{{\alpha ^{n - 1}} - {\beta ^{n - 1}}} \over {\alpha - \beta }} = {\alpha ^{n - 2}} + {\alpha ^{n - 3}}\beta + ... + {\beta ^{n - 2}}} \right]$$

$$ \Rightarrow $$ $${a_{n + 1}} + {a_{n - 1}} = {\alpha ^n} + {\beta ^n} = {b_n},\,\forall n \ge 1$$

So, option (b) is correct.

Now, $$\sum\limits_{n = 1}^\infty {{{{b_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} + {\beta ^n}} \over {{{10}^n}}}} $$

[as, b_n = $$\alpha $$ⁿ + $$\beta $$ⁿ]

= $$\sum\limits_{n = 1}^\infty {{{\left( {{\alpha \over {10}}} \right)}^n}} + \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} $$

$$ \because $$ $$\left[ {\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$$

$$ = {{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} + {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}} = {\alpha \over {10 - \alpha }} + {\beta \over {10 - \beta }}$$

$$ = {{10\alpha - \alpha \beta + 10\beta - \alpha \beta } \over {(10 - \alpha )(10 - \beta )}}$$

$$ = {{10(\alpha + \beta ) - 2\alpha \beta } \over {100 - 10(\alpha + \beta ) + \alpha \beta }}$$

$$ = {{10(1) - 2( - 1)} \over {100 - 10(1) - 1}}$$

[as $$\alpha $$ + $$\beta $$ = 1 and $$\alpha $$$$\beta $$ = $$ - $$1]

$$ = {{12} \over {89}}$$

So, option (a) is not correct.

$$ \because $$ $${\alpha ^2} = \alpha + 1$$ and $${\beta ^2} = \beta + 1$$

$$ \Rightarrow $$$${\alpha ^{n + 2}} = {\alpha ^{n + 1}} + {\alpha ^n}$$ and $${\beta ^{n + 2}} = {\beta ^{n + 1}} + {\beta ^n}$$

$$ \Rightarrow $$ $$({\alpha ^{n + 2}} + {\beta ^{n + 2}}) = ({\alpha ^{n + 1}} + {\beta ^{n + 1}}) + ({\alpha ^n} + {\beta ^n})$$

$$ \Rightarrow {a_{n + 2}} = {a_{n + 1}} + {a_n}$$

Similarly, $${a_{n + 1}} = {a_n} + {a_{n - 1}}$$

$${a_n} = {a_{n - 1}} + {a_{n - 2}}$$

..............

............

$${a_3} = {a_2} + {a_1}$$

On adding, we get

$${a_{n + 2}} = ({a_n} + {a_{n - 1}} + {a_{n - 2}} + ... + {a_2} + {a_1}) + {a_2}$$

$$ \because $$ $$\left[ {{a_2} = {{{\alpha ^2} - {\beta ^2}} \over {\alpha - \beta }} = \alpha + \beta = 1} \right]$$

So, $${a_{n + 2}} - 1 = {a_1} + {a_2} + {a_3} + ...... + {a_n}$$

So, option (c) is also correct.

And, now $$\sum\limits_{n = 1}^\infty {{{{a_n}} \over {{{10}^n}}}} = \sum\limits_{n = 1}^\infty {{{{\alpha ^n} - {\beta ^n}} \over {(\alpha - \beta ){{10}^n}}}} $$

$$ = {1 \over {\alpha - \beta }}\left[ {\sum\limits_{n = 1}^\infty {{{\left( {{a \over {10}}} \right)}^n}} - \sum\limits_{n = 1}^\infty {{{\left( {{\beta \over {10}}} \right)}^n}} } \right]$$

$$ = {1 \over {\alpha - \beta }}\left[ {{{{\alpha \over {10}}} \over {1 - {\alpha \over {10}}}} - {{{\beta \over {10}}} \over {1 - {\beta \over {10}}}}} \right]$$,

$$\left[ {as\left| {{\alpha \over {10}}} \right| < 1\,and\,\left| {{\beta \over {10}}} \right| < 1} \right]$$

$$ = {{10(\alpha - \beta )} \over {(\alpha - \beta )[100 - 10(\alpha + \beta ) + \alpha \beta ]}}$$

$$ = {{10} \over {100 - 10 - 1}} = {{10} \over {89}}$$

Hence, options (b), (c) and (d) are correct.