Advertisements
Advertisements
प्रश्न
Find the middle terms in the expansion of
`(3x + x^2/2)^8`
Advertisements
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
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The middle term in the expansion of `(x + 1/x)^10` is
Expand `(2x^2 - 3/x)^3`
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 994
Compute 97
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
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
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
