Advertisements
Advertisements
Question
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Advertisements
Solution
General term Tr+1 = `""^10"C"_"r" (x^2)^(10-"r") (1/x^3)^"r"`
= `""^10"C"_"r" x^(20-2"r") 1/(x^3"r")`
= `""^10"C"_"r" x^(20 - 2"r")* x^(-3"r")`
= `""^10"C"_"r" x^(20 - 5"r")`
To find coefficient of x15 we have to equate x power to 15
i.e. 20 – 5r = 15
20 – 15 = 5r
⇒ 5r = 5
⇒ r = `5/5` = 1
So the coefficient of x15 is 10C1 = 10
APPEARS IN
RELATED QUESTIONS
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
(2a – 3b)4
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 term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
Sum of the binomial coefficients is
Compute 1024
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 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]
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`
