Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
(2a – 3b)4
Find the Co-efficient of x11 in the expansion of `(x + 2/x^2)^17`
The middle term in the expansion of `(x + 1/x)^10` is
Expand `(2x^2 - 3/x)^3`
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
Find the last two digits of the number 3600
In the binomial expansion of (a + b)n, if the coefficients of the 4th and 13th terms are equal then, find n
If the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n
