Advertisements
Advertisements
Question
Expand `(2x^2 - 3/x)^3`
Advertisements
Solution
`(2x^2 - 3/x)^3 = ""^3"C"_0 (2x^2)^3 + ""^3"C"_1 (2x^2)^2 (- 3/x) + ""^3"C"_2 (2x^2)^1 (- 3/x)^2 + ""^3"C"_3 (- 3/x)^3`
3C0 = 3C3 = 1 ; 3C1 = 3C2 = 3
= `1(8)(x^6) + 3(4x^4) (-3/x) + 3(2x^2) (9/x^2) + 1(- 27/x^3)`
= `8x^6 - (36x^4)/x + (54x^2)/x^2 - 27/x^3`
= `8x^6 - 36x^3 + 54 - 27/x^3`
APPEARS IN
RELATED QUESTIONS
Expand the following by using binomial theorem.
`(x + 1/y)^7`
Find the 5th term in the expansion of (x – 2y)13.
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
Prove that the term independent of x in the expansion of `(x + 1/x)^(2n)` is `(1*3*5...(2n - 1)2^n)/(n!)`.
Find the Co-efficient of x11 in the expansion of `(x + 2/x^2)^17`
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 the binomial coefficients is
Compute 994
Compute 97
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
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
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
If n is a positive integer and r is a non-negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal
If a and b are distinct integers, prove that a − b is a factor of an − bn, whenever n is a positive integer. [Hint: write an = (a − b + b)n and expaand]
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 value of 2 + 4 + 6 + … + 2n is
