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Composite Function

Let a, b be t...

Question

Let $a, b$ be the roots of the quadratic equation $x^{2} + p x + 1 = 0,$ where $p^{2} = 1$ .
If $k$ is a positive integer, then $k \sum n = 1 (a^{2^{n}} + b^{2^{n}})$ is

0

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k

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- k

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0

k

is even and

- 1

k

is odd.

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Solution

The correct option is C $- k$
By the given quadratic equation, $a b = 1$ and $a + b = \pm 1$
$k \sum n = 1 (a^{2^{n}} + b^{2^{n}}) = a^{2} + b^{2} + a^{4} + b^{4} + \dots + (a^{2^{k}} + b^{2^{k}})$

Now, $a^{2} + b^{2} = (a + b)^{2} - 2 a b = 1 - 2 = - 1$
$a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 a^{2} b^{2} = 1 - 2 = - 1$
and so on.
Each term $(a^{2^{n}} + b^{2^{n}})$ in that summation is equal to $- 1$ .
Hence, $k \sum n = 1 (a^{2^{n}} + b^{2^{n}}) = - k$

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Let a,b be the roots of the quadratic equation x2+px+1=0, where p2=1. If k is a positive integer, then k∑n=1(a2n+b2n) is

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