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Find the term independent of x, x ≠ 0, in the expansion of (3x22-13x)15 - Mathematics

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Question

Find the term independent of x, x ≠ 0, in the expansion of `((3x^2)/2 - 1/(3x))^15`

Sum
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Solution

General Term `"T"_(r + 1) = ""^n"C"_r x^(n - r) y^r`

= `""^15"C"_r ((3x^2)/2)^(15 - r)  (- 1/(3x))^r`

= `""^15"C"_r (3/2)^(15 - r) * (x)^(30 - 2r) * (- 1/3)^r * 1/x^r`

= `""^15"C"_r (3/2)^(15 - r) * (x)^(30 - 2r - r) (-1)^r * 1/(3)^r`

= `""^15"C"_r (3/2)^(15 - r) * x^(30 - 3r) (-1)^r * 1/(3)^r`

For getting the term independent of x

30 – 3r = 0

⇒ r = 10

On putting the value of r in the above expression, we get

= `""^15"C"_10 (3/2)^(15 - 10) (-1)^10 * 1/(3)^10`

= `""^15"C"_10  (3)^5/(2)^5 * 1/(3)^10`

= `""^15"C"_10 * 1/((2)^5 * (3)^5)`

= `""^15"C"_10 (1/6)^5`

Hence, the required term = `""^15"C"_10 (1/16)^5`

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Chapter 8: Binomial Theorem - Exercise [Page 142]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 8 Binomial Theorem
Exercise | Q 1 | Page 142

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