Advertisements
Advertisements
Question
Prove that the term independent of x in the expansion of `(x + 1/x)^(2n)` is `(1*3*5...(2n - 1)2^n)/(n!)`.
Advertisements
Solution
There are (2n + 1) terms in the expansion.
∴ tn+1 is the middle term.
`"t"_(n+1) = 2_n"C"_n (x)^(2n - n) (1/x)^n`
`"t"_(n+1) = 2_n"C"_n x^n 1/x^n = 2n"C"_n (|__2n)/(|__n|__n)`
`= ((2n)(2n - 1)(2n - 2)...4*3*2*1)/(n(n-1)... 2 *1 |__n)`
`= ((2n - 1)(2n - 3)...3*1n(n - 1)(n - 2)...3*2*1 2^n)/(n(n-1)(n-2)...3*2*1|__n)`
`= ((2n - 1)...3*1)/(|__n) 2^n`
`= (1*3*5...(2n - 1))/(|__n)2^n`
APPEARS IN
RELATED QUESTIONS
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The last term in the expansion of (3 + √2 )8 is:
Sum of the binomial coefficients is
Compute 1024
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
If n is a positive integer and r is a non-negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal
