# If xp occurs in the expansion of (x2+1x)2n, prove that its coefficient is 2n!(4n-p3)!(2n+p3)! - Mathematics

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Sum

If xp occurs in the expansion of (x^2 + 1/x)^(2n), prove that its coefficient is (2n!)/(((4n - p)/3)!((2n + p)/3)!)

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#### Solution

Given expression is (x^2 + 1/x)^(2n)

General terms, "T"_(r + 1) = ""^n"C"_rx^(n - r) y^r

= ""^(2n)"C"_r (x^2)^(2n - r) * (1/x)^r

= ""^(2n)"C"_r (x)^(4n - 2r) * 1/x^r

= ""^(2n)"C"_r (x)^(4n - 2r - r)

= ""^(2n)"C"_r(x)^(4n - 3r)

If xp occurs in (x^2 + 1/x)^(2n)

Then 4n – 3r = p

⇒ 3r = 4n – p

⇒ r = (4n - p)/3

∴ Coefficient of xp = ""^(2n)"C"_r = ""^(""2n)"C"_((4n - p)/3)

= ((2n)!)/(((4n - p)/3)!(2n - (4n - p)/3)!)

= ((2n)!)/(((4n - p)/3)!((6n - 4n + p)/3)!)

= ((2n)!)/(((4n - p)/3)!((2n + p)/3)!)

Hence, the coefficient of xp = ((2n)!)/(((4n - p)/3)!((2n + p)/3)!)

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Chapter 8: Binomial Theorem - Exercise [Page 144]

#### APPEARS IN

NCERT Exemplar Class 11 Mathematics
Chapter 8 Binomial Theorem
Exercise | Q 16 | Page 144

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