Advertisements
Advertisements
प्रश्न
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
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
Expand the following by using binomial theorem.
(2a – 3b)4
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
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`
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`
Sum of the binomial coefficients is
Expand `(2x^2 -3sqrt(1 - x^2))^4 + (2x^2 + 3sqrt(1 - x^2))^4`
Compute 994
Compute 97
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
Find the last two digits of the number 3600
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
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
