If x2-2 x-2-x-12018-a2018 x2018-a2017 x2017-a1 x-a0-b1-x-1-b2-x-12-b2018-x-12 n 8- x - R-1 then ai- bi- c o n s t a n t

The correct options are A ∑^2018_i=0 a_i + ∑^2018_i=0b_i/2^i = ( J/2 )^2018 B ∑^2018_i=1b_i = 2^2018 (^2018C_1009)/2 D a_0 = ^2018C_1009 + 2^2018/2( x+1+1/x+1 ...

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Byju's Answer

Standard XII

Mathematics

General Tips for Choosing Function

If x2+2 x+2/x...

Question

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

\sum_{i = 0}^{2018} a_{i} + \sum_{i = 0}^{2018} \frac{b_{i}}{2^{i}} = {(\frac{J}{2})}^{2018}

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\sum_{i = 1}^{2018} b_{i} = \frac{2^{2018} - (^{2018} C_{1009})}{2}

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\sum_{i = 1}^{2018} b_{i} = 2^{2018} +^{2018} C_{1009}

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a_{0} = \frac{^{2018} C_{1009} + 2^{2018}}{2}

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Solution

The correct options are
A $\sum_{i = 0}^{2018} a_{i} + \sum_{i = 0}^{2018} \frac{b_{i}}{2^{i}} = {(\frac{J}{2})}^{2018}$
B $\sum_{i = 1}^{2018} b_{i} = \frac{2^{2018} - (^{2018} C_{1009})}{2}$
D $a_{0} = \frac{^{2018} C_{1009} + 2^{2018}}{2}$
${(x + 1 + \frac{1}{x + 1})}^{2018} =^{2018} C_{0} (x + 1)^{2018} +^{2018} C_{1} (x + 1)^{2016} + \dots \dots +^{2018} C_{1009} + \dots \dots^{2018} C_{2018} {(\frac{1}{x + 1})}^{2018}$
$= a_{2018} x^{2018} + \dots \dots + a_{1} x + a_{0} + \frac{b_{1}}{x + 1} + \dots \dots + \frac{b_{2018}}{(x + 1)^{2018}}$
Put $x = 1$
$\sum_{i = 0}^{2018} a_{i} + \sum_{i = 1}^{2018} \frac{b_{i}}{2^{i}} = {(\frac{5}{2})}^{2018}$
Put $x = 0$
$2^{2018} = (^{2018} C_{0} + \dots \dots +^{2018} C_{1009}) + ⎛ ⎜ ⎜ ⎜ ⎝^{2018} C_{1010} + \dots \dots +^{2018} C_{2018}      \sum_{i = 1}^{2018} b_{i} ⎞ ⎟ ⎟ ⎟ ⎠$
$\sum b_{i} = \frac{2^{2018} -^{2018} C_{1009}}{2}$
$a_{0} = \frac{2^{2018} +^{2018} C_{1009}}{2}$

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If (x2+2x+2x+1)2018=a2018x2018+a2017x2017+……+a1x+a0+b1x+1+b2(x+1)2+……+b2018(x+1)2018∀xϵR−{−1}then(ai, biϵconstant)

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