Advertisements
Advertisements
प्रश्न
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
उत्तर
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
संबंधित प्रश्न
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Find the term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
Find the Co-efficient of x11 in the expansion of `(x + 2/x^2)^17`
The last term in the expansion of (3 + √2 )8 is:
Compute 1024
Compute 994
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^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
