If α and β are the roots of the equation x2−2x+4=0, such that αn+βn=2kcosnπ3, then value of k is
A
n−1
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B
n
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C
n+1
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D
2n
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Solution
The correct option is Cn+1 x2−2x+4=0 ⇒x=2±√4−162=1±i√3
Let α=1+i√3=2[cosπ3+isinπ3]
and β=1−i√3=2[cosπ3−isinπ3]
So, αn+βn=2n[cosπ3+isinπ3]n+2n[cosπ3−isinπ3]n =2n[cosnπ3+isinnπ3+cosnπ3−isinnπ3] =2n⋅2cosnπ3=2n+1⋅cosnπ3 ∴k=n+1