(1)^any number = 1 ( 1)^odd number = 1 ( 1)^even number = +1 (1)^ a = 1/1^a=+1 ∴ nbsp;( 1)^2017 gives 1 which is less than nbsp;1. ...

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Null Matrix

12018 >

Question

$1^{2018}$ >

(- 1)^{2017}

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1^{- 2018}

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1^{2017}

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(- 1)^{2018}

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Solution

The correct option is A $(- 1)^{2017}$
${(1)}^{a n y n u m b e r} = 1$
${(- 1)}^{o d d n u m b e r} = - 1$
${(- 1)}^{e v e n n u m b e r} = + 1$
${(1)}^{- a} = \frac{1}{1^{a}} = + 1$
$∴$ $(- 1)^{2017}$ gives -1 which is less than 1.

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1^{2018}

{(\frac{x^{2} + 2 x + 2}{x + 1})}^{2018} = a_{2018} x^{2018} + a_{2017} x^{2017} + \dots \dots + a_{1} x + a_{0} + \frac{b_{1}}{x + 1} + \frac{b_{2}}{(x + 1)^{2}} + \dots \dots + \frac{b_{2018}}{(x + 1)^{2018}} \forall x ϵ R - {- 1} t h e n (a_{i}, b_{i} ϵ constant)

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