Maharashtra State BoardHSC Science (General) 11th
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Find the constant term (term independent of x) in the expansion of (2x+13x2)9 - Mathematics and Statistics

Sum

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

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Solution

Let tr+1 be the constant term in the expansion of `(2x + 1/(3x^2))^9`.

We know that, in the expansion of (a + b)n,

tr+1 = nCr an–r br 

Here a = 2x, b = `1/(3x^2)`, n = 9

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

= `""^9"C"_"r" 2^(9 - "r")*x^(9 - "r")*(1/3)^"r"*x^(-2"r")`

= `""^9"C"_"r" 2^(9 - "r")*(1/3)^"r"*x^(9 - 3"r")`

But tr+1 is a constant term

∴ power of x = 0

∴ 9 – 3r = 0

∴ r = 3

∴ the constant term = `""^9"C"_3*2^(9 - 3)*(1/3)^3`

= `(9 xx 8 xx 7)/(1 xx 2  xx 3) xx 2^6 xx 1/3^3`

= `84 xx 64 xx 1/27`

= `1792/9`

Concept: General Term in Expansion of (a + b)n
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APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.3 | Q 3. (i) | Page 80
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