Advertisements
Advertisements
प्रश्न
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
Advertisements
उत्तर
`(x - 2/x^2)^15 = (x + (-2)/x^2)^15` compare with the (x + a)n
Here x is x, a is `(-2)/x^2`, n = 15
Let the independent term of x occurs in the general term
`"t"_(r+1) = n"C"_r x^(n-r) a^r`
`"t"_(r+1) = 15"C"_r x^(15-r)((-2)/x^2)^r = 15"C"_r x^(15-r) (-2)^r/(x^2)^r`
`= 15"C"_r x^(15-r) (-2)^r/(x^(2r))`
`= 15"C"_r x^(15-r) * x^(-2r) (-2)^r = 15"C"_r x^(15-r-2r) * (-2)^r`
`= 15"C"_r x^(15-3r) * (-2)^r`
Independent term occurs only when x power is zero.
15 – 3r = 0
15 = 3r
r = 5
Using r = 5 in (1) we get the independent term
= 15C5 x0 (-2)5 [∵ (-2)5 = (-1)5 25 = -25]
= -32(15C5)
APPEARS IN
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(101)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
`(2x^2 + 1/x)^12`
The middle term in the expansion of `(x + 1/x)^10` is
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Compute 97
Find the coefficient of x15 in `(x^2 + 1/x^3)^10`
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
