Advertisements
Advertisements
प्रश्न
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Advertisements
उत्तर
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
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(101)4
Evaluate the following using binomial theorem:
(999)5
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
`(x + 1/x)^11`
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
Prove that the term independent of x in the expansion of `(x + 1/x)^(2n)` is `(1*3*5...(2n - 1)2^n)/(n!)`.
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
The last term in the expansion of (3 + √2 )8 is:
Expand `(2x^2 - 3/x)^3`
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 1024
Find the last two digits of the number 3600
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
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
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
