Advertisements
Advertisements
प्रश्न
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
Advertisements
उत्तर
`(2x^2 + 1/x)^12` Compare with the (x + a)n.
Here x is 2x2, a is `1/x`, n = 12.
Let the independent term of x occurs in the general term.
tr+1 = nCr xn-r ar
`"t"_(r+1) = 12"C"_r (2x^2)^(12-r) (1/x)^r = 12"C"_r 2^(12-r) x^(2(12-r)) x^-r`
= 12Cr 212-r x24-2r x-r
`= 12"C"_r 12^(12-r) x^(24-3r)`
Independent term occurs only when x power is zero
24 – 3r = 0
24 = 3r
r = 8
Put r = 8 in (1) we get the independent term as
= 12C8 212-8 x0
= 12C4 × 24 × 1
= 7920
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.
(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 - 2/(3x))^9`
Sum of the binomial coefficients is
Compute 994
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
