Advertisements
Advertisements
Question
Find the constant term of `(2x^3 - 1/(3x^2))^5`
Advertisements
Solution
General term Tr+1 = `""^5"C"_"r" (2x^3)^(5 - "r") ((-1)/(3x^2))^"r"`
= `""^5"C"_"r" 2^(5 - "r") x^(15 - 3"r") (- 1)^"r" 1/(3^"r") 1/(x^(2"r"))`
= `""^15"C"_"r" (2^(5 - "r"))/(3^"r") (- 1)^"r" x^(15 - 3"r" - 2"r")`
= `""^5"C"_"r" (- 1)^"r" (2^(5 - "r"))/(3^"r") x^(15 - 5"r")`
To find the constant term
15 – 5r = 0
⇒ 5r = 15
⇒ r = 3
∴ Constant term = `""^5"C"_3 (- 1)^3 (2^(5 - 3))/3^3`
= `(5 xx 4 xx 3)/(3 xx 2 xx 1) (- 1) ((2^2))/3^3`
= `(10 - (- 1)(4))/27`
= `(- 40)/27`
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
`(3x + x^2/2)^8`
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
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!)`
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
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 1024
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
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 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 (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
