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Find the term independent of x in the expansion of (x2-23x)9 - Business Mathematics and Statistics

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प्रश्न

Find the term independent of x in the expansion of

`(x^2 - 2/(3x))^9`

योग

उत्तर

Let the independent form of x occurs in the general term, tr+1 = nCr xn-r ar

Here x is x2, a is `(-2)/(3x)` and n = 9

∴ `"t"_(r+1) = 9"C"_"r" (x^2)^(9-r) ((-2)/(3x))^r = 9"C"_r  x^(2(9-r)) ((-2)^r/(3^rx^r))`

`= 9"C"_r  x^(18-2r) * x^(-r) (-2)^r/3^r`

`= 9"C"_r  x^(18-2r-r) (-2)^r/3^r = 9"C"_r  x^(18-3r)  (-2)^r/3^r`

Independent term occurs only when x power is zero.

18 – 3r = 0

⇒ 18 = 3r

⇒ r = 6

Put r = 6 in (1) we get the independent term as 9C6 x0 `(-2)^6/3^6`

`= 9"C"_3 (2/3)^6`   ..[∵ 9C6 = 9C9-6 = 9C3]

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Binomial Theorem
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Algebra - Exercise 2.6 [पृष्ठ ४५]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
अध्याय 2 Algebra
Exercise 2.6 | Q 5. (i) | पृष्ठ ४५
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