# Find the term independent of x in the expansion of (1 + x + 2x3) (32x2-13x)9 - Mathematics

Sum

Find the term independent of x in the expansion of (1 + x + 2x3) (3/2 x^2 - 1/(3x))^9

#### 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

Concept: General and Middle Terms
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Chapter 8: Binomial Theorem - Exercise [Page 144]

#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 17 | Page 144

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