Advertisements
Advertisements
Question
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
Solution
General term Tr+1 = `""^6"C"_"r" (x^2)^(6 - "r") ((-1)/x^3)^"r"`
= `""^6"C"_"r" x^(12 - 2"r") (- 1)^"r" 1/(x^(3"r"))`
= `""^6"C"_"r" (- 1)^"r" x^(12 - 2"r" - 3"r")`
= `""^16"C"_"r" ( 1)^"r"x^(12 - 5"r")`
To find coefficient of x6
12 – 5r = 6
12 – 6 = 5r
⇒ 5r = 6
⇒ r = `6/5` which is not an integer.
∴ There is no term involving x6.
To find coefficient of x2
12 – 5r = 2
5r = 12 – 2 = 10
⇒ r = 2
So coefficient of x2 is `""^6"C"_2 (- 1)^2 = (6 xx 5)/(2 xx 1)(1)` = 15
APPEARS IN
RELATED QUESTIONS
Evaluate the following using binomial theorem:
(999)5
Find the middle terms in the expansion of
`(x + 1/x)^11`
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
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:
Sum of the binomial coefficients is
Expand `(2x^2 - 3/x)^3`
Compute 1024
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
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 the binomial coefficients of three consecutive terms in the expansion of (a + x)n are in the ratio 1 : 7 : 42, then find n
Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
