Question

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IIT Based Question!


Plz, solve the Question in attachment.

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Answer 1

Yep! the question of IIT Main.

Answer:

→ 1 .

Step-by-step explanation:

Given :

→ $\alpha$ and $\beta$ are the roots of equation : x² - x - 1 = 0 .

Then,

$\sf { \alpha }^{2} - \alpha - 1 = 0. \\ \\ \implies \sf { \alpha }^{2} - 1 = \alpha .$

$\sf { \beta }^{2} - \beta - 1 = 0. \\ \\ \implies \sf { \beta }^{2} - 1 = \beta .$

$\sf \because \frac{a_{2012} - a_{2010}}{a_{2011}} \\ \\ \sf = \frac{ \frac{( { \alpha }^{2012} - { \beta }^{2012}) }{ \alpha - \beta } - \frac{( { \alpha }^{2010} - { \beta }^{2010} ) }{ \alpha - \beta } }{ \frac{ { \alpha }^{2011} - { \beta }^{2011} }{ \alpha - \beta } } . \\ \\ \sf = \frac{ \frac{( { \alpha }^{2012} - { \beta }^{2012} - { \alpha }^{2010} - { \beta }^{2010}) }{ \cancel{ (\alpha - \beta )}} }{ \frac{ { \alpha }^{2011} - { \beta }^{2011} }{ \cancel{( \alpha - \beta )}} } . \\ \\ \sf = \frac{ ( { \alpha }^{2012} - { \beta }^{2012} - { \alpha }^{2010} - { \beta }^{2010}) }{ { \alpha }^{2011} - { \beta }^{2011} }. \\ \\ \sf = \frac{ { \alpha }^{2010}( { \alpha }^{2} - 1) - { \beta }^{2010} ( { \beta }^{2} - 1) }{ { \alpha }^{2011} - { \beta }^{2011} } . \\ \\ \sf = \frac{ { \alpha }^{2010}( \alpha ) - { \beta }^{2010}( \beta ) }{ { \alpha }^{2011} - { \beta }^{2011} } . \\ \\ \sf = \frac{ { \alpha }^{2011} - { \beta }^{2011} }{ { \alpha }^{2011} - { \beta }^{2011} } . \\ \\ \huge{ \pink{ \boxed{ \tt = 1.}}}$

Hence, it is solved.

Answer 2

Step-by-step explanation:

[refer the attachments]

first of all find the roots of a given equation α and β ,then put it into the an .

Now, we have to find

[a2012 - a2010]/a2011 , So substitute the values of a2012 , a2010 , a2011 with the help of (an). Now simply do the calculations using indices rules and BODMAS property .

Hence we get the final value 1.