Maharashtra State BoardHSC Science (General) 11th
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Expand: (2x-1x)6 - Mathematics and Statistics

Sum

Expand: `(2x - 1/x)^6`

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Solution

`(2x - 1/x)^6 = ""^6"C"_0(2x)^6 - ""^6"C"_1(2x)^5(1/x) + ""^6"C"_2(2x)^4(1/x)^2 - ""^6"C"_3(2x)^3(1/x)^2 + ""^6"C"_4(2x)^2(1/x)^4 - ""^6"C"_5(2x)(1/x)^5 + ""^6"C"_6(1/x)^6`

Now, 6C0 = 1 = 6C6

6C1 = 6 = 6C4 

6C2 = `(6 xx 5)/(1 xx 2)` = 15 = 6C4

6C3 = `(6 xx 5 xx 4)/(1 xx 2 xx 3)` = 20

∴ `(2x - 1/x)^6 = 1 xx 64x^6 - 6 xx 32x^5 xx 1/ x + 15 xx 16x^4 xx 1/x^2 - 20 xx 8x^3 xx 1/ x^3 + 15 xx 4x^2 xx 1/ x^4 - 6 xx 2x xx 1/x^5 + 1 xx 1/x^6` 

= `64x^6 - 192x^4 + 240x^2 - 160 + 60/x^2 - 12/x^4 + 1/x^6`.

Concept: Binomial Theorem for Positive Integral Index
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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 4.2 | Q 2. (ii) | Page 77
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