If (x+1/x+1)^(6)=a(0)+(a(1)x+(b(i))/(x))...

If `(x+1/x+1)^(6)=a_(0)+(a_(1)x+(b_(i))/(x))+(a_(2)x^(2)+(b_(2))/(x^(2)))+"...."+(a_(6)x^(6)+(b_(6))/(x^(6)))`, then

`a_() = 141`

`a_(5) = 6`

`underset(i=1)overset(6)sum a_(i) + b_(i) = 588`

`underset(i=1)overset(6)suma_(i) + b_(i) = 3^(6)`

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The correct Answer is:

A, B, C

`(x+1/x+1)^(6)=underset(r=0)overset(n)sum.^(6)C_(r)(x+1/x)^(r)`
(1) `a_(0) = 1 +.^(6)C_(2)..^(2)C_(1) + .^(6)C_(4)+.^(4)C_(2)+.^(6)C_(6).^(6)C_(3)`
`= 1+30+90+20 = 141`
(2) `a_(5) = .^(6)C_(5) xx` Coefficient of `x^(5)` in `(x+1/x)^(5)`
`= .^(6)C_(5).^(5)C_(0) = 6`
(3) Putting , `x = 1`
`a_(0) + (a_(1) + b_(1)) + (a_(2)+b_(2))= 3^(6)`
`:. underset(i=1)overset(6)sum(a_(i)+b_(i)) = 3^(6) - 141 = 588`

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