Advertisements
Advertisements
प्रश्न
Find the term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
उत्तर
Let the independent form of x occurs in the general term, tr+1 = nCr xn-r ar
Here x is x2, a is `(-2)/(3x)` and n = 9
∴ `"t"_(r+1) = 9"C"_"r" (x^2)^(9-r) ((-2)/(3x))^r = 9"C"_r x^(2(9-r)) ((-2)^r/(3^rx^r))`
`= 9"C"_r x^(18-2r) * x^(-r) (-2)^r/3^r`
`= 9"C"_r x^(18-2r-r) (-2)^r/3^r = 9"C"_r x^(18-3r) (-2)^r/3^r`
Independent term occurs only when x power is zero.
18 – 3r = 0
⇒ 18 = 3r
⇒ r = 6
Put r = 6 in (1) we get the independent term as 9C6 x0 `(-2)^6/3^6`
`= 9"C"_3 (2/3)^6` ..[∵ 9C6 = 9C9-6 = 9C3]
APPEARS IN
संबंधित प्रश्न
Expand the following by using binomial theorem.
(2a – 3b)4
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
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!)`
Expand `(2x^2 - 3/x)^3`
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 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]