Anmol ᗷᖇO ᗯᕼEᖇE ᗩᖇE YOᑌ?¿
Answers
Answer:
don't know sry
good night
Step-by-step explanation:
Yep! the question of IIT Main.
Answer:
→ 1 .
Step-by-step explanation:
Given :
→ \alphaα and \betaβ are the roots of equation : x² - x - 1 = 0 .
Then,
\begin{gathered}\sf { \alpha }^{2} - \alpha - 1 = 0. \\ \\ \implies \sf { \alpha }^{2} - 1 = \alpha .\end{gathered}
α
2
−α−1=0.
⟹α
2
−1=α.
\begin{gathered}\sf { \beta }^{2} - \beta - 1 = 0. \\ \\ \implies \sf { \beta }^{2} - 1 = \beta .\end{gathered}
β
2
−β−1=0.
⟹β
2
−1=β.
\begin{gathered}\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.}}}\end{gathered}
∵
a
2011
a
2012
−a
2010
=
α−β
α
2011
−β
2011
α−β
(α
2012
−β
2012
)
−
α−β
(α
2010
−β
2010
)
.
=
(α−β)
α
2011
−β
2011
(α−β)
(α
2012
−β
2012
−α
2010
−β
2010
)
.
=
α
2011
−β
2011
(α
2012
−β
2012
−α
2010
−β
2010
)
.
=
α
2011
−β
2011
α
2010
(α
2
−1)−β
2010
(β
2
−1)
.
=
α
2011
−β
2011
α
2010
(α)−β
2010
(β)
.
=
α
2011
−β
2011
α
2011
−β
2011
.
=1.