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Maths-

General

Easy

Question

$open square brackets 5 square root of 5 plus 11 close square brackets to the power of 2 n plus 1 end exponent$ is always

an odd integer
an even integer
0
none of these.

The correct answer is: an even integer

Let $open parentheses 5 square root of 5 plus 11 close parentheses to the power of 2 n plus 1 end exponent equals I plus f$ , where I is an integer and 0 $less or equal than$ f < 1
Let $open parentheses 5 square root of 5 minus 11 close parentheses to the power of 2 n plus 1 end exponent equals f ´$ , where 0 < f ’ < 1
$therefore I plus f minus f ´ equals open parentheses 5 square root of 5 plus 11 close parentheses to the power of 2 n plus 1 end exponent minus open parentheses 5 square root of 5 minus 11 close parentheses to the power of 2 n plus 1 end exponent$
$equals 2 open square brackets 2 n plus 1 C subscript 1 end subscript open parentheses 5 square root of 5 close parentheses to the power of 2 n end exponent open parentheses 11 close parentheses plus to the power of 2 n plus 1 end exponent C subscript 3 end subscript open parentheses 5 square root of 5 close parentheses to the power of 2 n minus 2 end exponent open parentheses 11 close parentheses to the power of 3 end exponent plus....... close square brackets$
=2 (An integer) = 2k where k $element of$ N
= An even integer.
Hence f – f ’ = 2k – I = An integer.
But – 1 < f – f ’ < 1 \ 0 = 2k – I
\ I = 2k = an even integer.
Hence integral part of $open parentheses 5 square root of 5 plus 11 close parentheses to the power of 2 n plus 1 end exponent$ is an even integer.

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