Maharashtra State BoardHSC Science (General) 11th
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Answer the following: Expand (2x3-32x)4 - Mathematics and Statistics

Sum

Answer the following:

Expand `((2x)/3 - 3/(2x))^4`

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Solution

`((2x)/3 - 3/(2x))^4 = ""^4"C"_0 ((2x)/3)^4 - ""^4"C"_1 ((2x)/3)^3 (3/(2x)) + ""^4"C"_2 ((2x)/3)^2 (3/(2x))^2 - ""^4"C"_3 ((2x)/3)(3/(2x))^3 + ""^4"C"_4 (3/(2x))^4`

Now, 4C0 = 1 = 4C4

4C1 = 4 = 4C3

4C2 = `(4 xx 3)/(1 xx 2)` = 6

∴ `((2x)/3 - 3/(2x))^4 = 1 xx (16x^4)/81 - 4 xx (8x^3)/27 xx 3/(2x) + 6 xx (4x^2)/9 xx 9/(4x^2) - 4 xx (2x)/3 xx 27/(8x^3) + 1 xx 81/(16x^4)`

= `(16x^4)/81 - (16x^2)/9 + 6 - 9/x^2 + 81/(16x^4)`

Concept: Binomial Theorem for Negative Index Or Fraction
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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. (5) | Page 85
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