Advertisements
Advertisements
Question
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
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:
(101)4
Expand the following by using binomial theorem.
(2a – 3b)4
The last term in the expansion of (3 + √2 )8 is:
Expand `(2x^2 - 3/x)^3`
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Find the constant term of `(2x^3 - 1/(3x^2))^5`
Find the last two digits of the number 3600
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 (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 value of 2 + 4 + 6 + … + 2n is
