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:
(101)4
Expand the following by using binomial theorem.
(2a – 3b)4
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
The constant term in the expansion of `(x + 2/x)^6` is
The last term in the expansion of (3 + √2 )8 is:
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Find the last two digits of the number 3600
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
