Advertisements
Advertisements
Question
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Advertisements
Solution
Here x is 3x, a is `x^2/2`, n = 8, which is even.
∴ The only one middle term = `("t"_(n+1))/2 = ("t"_(8+1))/2 = "t"_5`
General term tr+1 = nCr xn-r ar
`"t"_5 = "t"_(4+1) = 8"C"_4 (3x)^(8-4) (x^2/2)^4 = 8"C"_4 (3x)^4 (x^2)^4/2^4`
`= 8"C"_4 3^4x^4 x^8/2^4 = 8"C"_4 3^4/2^4 x^12 = 8"C"_4 81/16 x^12`
APPEARS IN
RELATED QUESTIONS
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
(2a – 3b)4
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The constant term in the expansion of `(x + 2/x)^6` is
Sum of the binomial coefficients is
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
In the binomial expansion of (a + b)n, if the coefficients of the 4th and 13th terms are equal then, find n
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
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
