In the expansion of (1+ x + 7/x)^11find ...

In the expansion of `(1+ x + 7/x)^11`find the term not containing x.

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The correct Answer is:

C

`(1+x+7/x)^(11) = ((7+x+x^(2))/(x))^(11)`
So, we have to find the coefficient of `x^(11)` in `(7+x+x^(2))^(11)`
Now, `(7+x+x^(2))^(11)`
`=[((7+x)+x^(2))]^(11)`
`= .^(11)C_(0)(7+x)^(11)+.^(11)C_(1)(7+x)^(10)x^(2)+.^(11)C_(2)(7+x)^(9)x^(4)"....."+"...."`
`:.` Required coefficient `= .^(11)C_(0)+.^(11)C_(1)xx.^(10)C_(9)xx7+.^(11)C_(2)xx.^(9)C_(7)`
`xx7^(2)+.^(11)C_(3)xx.^(8)C_(5) xx7^(3)+.^(11)C_(4)`
`xx.^(7)C_(3)xx7^(4)+.^(11)C_(5)xx.^(6)C_(1)xx7^(5)`
`= underset(r=0)overset(5)sum.^(11)C_(r).^(11-r)C_(11-2r)7^(r)`

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