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Algebra Level 2

If a + 1 a = 2 a+\dfrac{1}{a}=2 , find the value of: a n + 1 a n \large a^n+\frac{1}{a^n}

where n = 2 k n=2^k ; k ∈ N k\in \mathbb N .

Can't be ditermined 0 1 2 ∞ \infty

Md Mehedi Hasan by Md Mehedi Hasan , 22, Bangladesh

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2 solutions

Given that a + 1 a = 2 a+\dfrac 1a = 2 . Multiplying a a throughout and rearrange a 2 − 2 a + 1 = 0 a^2 - 2a + 1 = 0 , ⟹ ( a − 1 ) 2 = 0 \implies (a-1)^2 = 0 ⟹ a = 1 \implies a = 1 . Therefore, a n + 1 a n = 1 + 1 = 2 a^n + \dfrac 1{a^n} = 1+1 = 2 for all n ∈ Z n \in \mathbb Z .

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Md Mehedi Hasan
Nov 5, 2017

a n + 1 a n = a 2 k + 1 a 2 k = ( a 2 k − 1 ) 2 + ( 1 a 2 k − 1 ) 2 = ( a 2 k − 1 + 1 a 2 k − 1 ) 2 − 2 = { ( a 2 k − 2 + 1 a 2 k − 2 ) 2 − 2 } 2 − 2 = { ( a 2 k − 2 + 1 a 2 k − 2 ) 2 − 2 } 2 − 2 = [ { ( a 2 k − k + 1 a 2 k − k ) 2 − 2 } 2 . . . . . . . . . ] 2 − 2 = [ { ( a + 1 a ) 2 − 2 } 2 . . . . . . . . . ] 2 − 2 = [ { ( 2 ) 2 − 2 } 2 . . . . . . . . . ] 2 − 2 = [ { 4 − 2 } 2 . . . . . . . . . ] 2 − 2 = 4 − 2 = 2 \large{a^{ n }+\frac { 1 }{ a^{ n } } \\ =a^{ 2^{ k } }+\frac { 1 }{ a^{ 2^{ k } } } \\ =(a^{ 2^{ k-1 } })^{ 2 }+\left( \frac { 1 }{ a^{ 2^{ k-1 } } } \right) ^{ 2 }\\ =\left( a^{ 2^{ k-1 } }+\frac { 1 }{ a^{ 2^{ k-1 } } } \right) ^{ 2 }-2\\ =\left\{ \left( a^{ 2^{ k-2 } }+\frac { 1 }{ a^{ 2^{ k-2 } } } \right) ^{ 2 }-2 \right\} ^{ 2 }-2\\ =\left\{ \left( a^{ 2^{ k-2 } }+\frac { 1 }{ a^{ 2^{ k-2 } } } \right) ^{ 2 }-2 \right\} ^{ 2 }-2\\ =\left[ \left\{ \left( a^{ 2^{ k-k } }+\frac { 1 }{ a^{ 2^{ k-k } } } \right) ^{ 2 }-2 \right\} ^{ 2 }......... \right] ^{ 2 }-2\\ =\left[ \left\{ \left( a+\frac { 1 }{ a } \right) ^{ 2 }-2 \right\} ^{ 2 }......... \right] ^{ 2 }-2\\ =\left[ \left\{ \left( 2 \right) ^{ 2 }-2 \right\} ^{ 2 }......... \right] ^{ 2 }-2\\ =\left[ \left\{ 4-2 \right\} ^{ 2 }......... \right] ^{ 2 }-2\\ =4-2\\ =\boxed { 2 } }

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I didn't understand your solution.

Munem Shahriar - 3 years, 7 months ago

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In 4th line, I use a 2 + b 2 = ( a + b ) 2 − 2 a b a^2+b^2=(a+b)^2-2ab formula. If I apply this formula in every line, one time 7th line come. Then I can input the value of a + b = 2 a+b=2 . Then every time this 2 2 will square and minus 2 2 and remain 2 2 . In this way, at last the result come 2 2

Md Mehedi Hasan - 3 years, 7 months ago

It's a long and hard way to solve the problem this way. The one of @Chew-Seong Cheong is way faster!

Peter van der Linden - 3 years, 7 months ago

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Yes, I agree. But it's also a general process! Thanks for comment.

Md Mehedi Hasan - 3 years, 7 months ago

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