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If Un=(sqrt(3)+1)^(2n)+(sqrt(3)-1)^(2n) ...

If `U_n=(sqrt(3)+1)^(2n)+(sqrt(3)-1)^(2n)` , then prove that `U_(n+1)=8U_n-4U_(n-1)dot`

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`U_(n) = [(sqrt(3) + 1)^(2)]^(n) + [(sqrt(3) - 1)^(2)]^(n)`
`= (4+2sqrt(3))^(n) + (4-2sqrt(3))^(n)`
` = alpha^(n) + beta^(n)` where `alpha + beta = 8, alphabeta = 4`
Now, `8Y_(n) = (alpha+beta)(alpha^(n)+beta^(n))`
`= alpha^(n+1)+beta^(n+1)+betaalpha^(n)`
`= U_(n+1)+alphabeta(alpha^(n-1)+beta^(n-1))`
`= U_(n+1)+4U_(n-1)`
`rArr U_(n+1) = 8U_(n) - 4U_(n-1)`.
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Knowledge Check

  • If U_n=2 cosntheta then U_1U_n-U_(n-1) is equal to

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  • If n is a possible integer, then (sqrt3+1)^(2n)-(sqrt3-1)^(2n) is

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