Maharashtra State BoardHSC Science (General) 11th
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Answer the following: Find the constant term in the expansion of (4x23+32x)9 - Mathematics and Statistics

Sum

Answer the following:

Find the constant term in the expansion of `((4x^2)/3 + 3/(2x))^9`

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Solution

Let tr+1 be the constant term in the expansion of `((4x^2)/3 + 3/(2x))^9`.

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

tr+1 = nCr, an–r b 

Here a = `(4x^2)/3`, b = `3/(2x)`, n = 9

∴ tr+1 = `""^9"C"_"r" ((4x^2)/3)^(9-"r") (3/(2x))^"r"`

= `""^9"C"_"r" (4/3)^(9-"r") x^(18-2"r")*(3/2)^"r" x^(-"r")`

= `""^9"C"_"r" (4/3)^(9 - "r") (3/2)^"r" x^(18 - 3"r")`

But tr+1 is a constant term

∴ power of x = 0

∴ 18 – 3r = 0

∴ r = 6

∴ the constant term

= `""^9"C"_6 (4/3)^(9 - 6) (3/2)^6`

= `""^9"C"_3 (4/3)^3 (3/2)^6`   ...[∵ nCr = nCn–r]

= `(9 xx 8 xx 7)/(1 xx 2 xx 3) xx 64/27 xx (27 xx 27)/64`

= 2268

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
Miscellaneous Exercise 4 | Q II. (10) (i) | Page 86
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