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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता ११

Find the constant term of (2x3-13x2)5

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प्रश्न

Find the constant term of `(2x^3 - 1/(3x^2))^5`

बेरीज
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उत्तर

General term Tr+1 = `""^5"C"_"r" (2x^3)^(5 - "r") ((-1)/(3x^2))^"r"`

= `""^5"C"_"r" 2^(5 - "r") x^(15 - 3"r") (- 1)^"r" 1/(3^"r") 1/(x^(2"r"))`

= `""^15"C"_"r" (2^(5 - "r"))/(3^"r") (- 1)^"r" x^(15 - 3"r" - 2"r")`

= `""^5"C"_"r" (- 1)^"r" (2^(5 - "r"))/(3^"r") x^(15 - 5"r")`

To find the constant term

15 – 5r = 0

⇒ 5r = 15

⇒ r = 3

∴ Constant term = `""^5"C"_3 (- 1)^3 (2^(5 - 3))/3^3`

= `(5 xx 4 xx 3)/(3 xx 2 xx 1) (- 1) ((2^2))/3^3`

= `(10 - (- 1)(4))/27` 

= `(- 40)/27` 

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Binomial Theorem
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Binomial Theorem, Sequences and Series - Exercise 5.1 [पृष्ठ २१०]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 5 Binomial Theorem, Sequences and Series
Exercise 5.1 | Q 7 | पृष्ठ २१०

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