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Question
Find the term independent of x in the expansion of (1 + x + 2x3) `(3/2 x^2 - 1/(3x))^9`
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Solution
Given expression is (1 + x + 2x3) `(3/2 x^2 - 1/(3x))^9`
Let us consider `(3/2 x^2 - 1/(3x))^9`
General Term `"T"_(r + 1) = ""^n"C"_r x^(n - r) y^r`
`"T"_(r + 1) = ""^9"C"_r (3/2 x^2)^(9 - r) (- 1/(3x))^r`
= `""^9"C"_r (3/2)^(9 - r) (x)^(18 - 2r) * (- 1/3)^r * 1/(x)^r`
= `""^9"C"_r (3/2)^(9 - r) (x)^(18 - 2r - r) * (- 1/3)^r`
= `""^9"C"_r (3/2)^(9 - r) (- 1/3)^r * x^(18 - 3r)`
So, the general term in the expansion of
`(1 + x + 2x^3) (3/2 x^2 - 1/(3x))^9`
= `""^9"C"_r (3/2)^(9 - r) (- 1/3)^r * (x)^(18 - 3r) + ""^9"C"_r (3/2)^(9 - r) (- 1/3)^r * (x)^(19 - 3r) + 2 * ""^9"C"_r (3/2)^(9 - r) (- 1/3)^r * (x)^(21 - 3r)`
For getting the term independent of x,
Put 18 – 3r = 0, 19 – 3r = 0 and 21 – 3r = 0, we get
r = 6
r = `19/3` and r = 7
The possible value of r are 6 and 7 ```.....(because r ≠ 19/3)`
∴ The term independent of x is
= `""^9"C"_6 (3/2)^(9 - 6) (- 1/3)^6 + 2 * ""^9"C"_7 (3/2)^(9 - 7) (- 1/3)^7`
= `(9 xx 8 xx 7 xx 6!)/(3 xx 2 xx 1 xx 6!) * 3^3/2^3 * 1/3^6 - 2 * (9 xx 8 xx 7!)/(7!2 xx 1) * 3^2/2^2 * 1/3^7`
= `84/8 * 1/3^3 - 36/4 * 2/3^5`
= `7/18 - 2/27`
= `(21 - 4)/54`
= `17/54`
Hence, the required term = `17/54`
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