Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If x exponent of n occurs in the expansion of (x^2+1/x)^2n then prove that its coefficient is (2n)!/(4n-p/3)!(2n+p/3)!

If xn  occurs expansion of \left (x^{n}+\frac{1}{x} \right )^{2n} then prove that its coefficient is \frac{\left ( 2n \right )!}{\left ( \frac{4n-p}{3} \right )!\left ( \frac{2n+p}{3} \right )!}

Answers (1)

Given: (x2 + 1/x)2n

Thus, Tr+1 =2nCr(x2)2n-r(1/x)r

                 = 2nCrx4n-3r

Now, if xn will occur in the expansion, then,

Let us consider 4n – 3r = p

Thus,

3r = 4n – p

r = 4n – p/3

Thus, coefficient of xp = 2nCr

                                               = (2n)!/r!(2n-r)!

                                               = \frac{\left ( 2n \right )!}{\left ( \frac{4n-p}{3} \right )!\left ( \frac{2n+p}{3} \right )!}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question