Maharashtra State BoardHSC Science (Electronics) 11th
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Expand: (5-2)5 - Mathematics and Statistics

Sum

Expand: `(sqrt(5) - sqrt(2))^5`

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Solution

`(sqrt(5) - sqrt(2))^5 = ""^5"C"_0(sqrt(5))^5 - 5"C"_1(sqrt(5))^4 (sqrt(2)) + ""^5"C"_2(sqrt(5))^3(sqrt(2))^2 - ""^5"C"_3(sqrt(5))^2(sqrt(2))^3 + ""^5"C"_4 (sqrt(5))(sqrt(2))^4 - ""^5"C"_5 (sqrt(2))^5`

Now, 5C0 = 1 = 5C5

5C1 = 5 = 5C4

5C2 = `(5 xx 4)/(1 xx 2)` = 10 = 5C3

∴ `(sqrt(5) - sqrt(2))^5 = 1 xx 25sqrt(5) - 5 xx 25 xx sqrt(2) + 10 xx 5sqrt(5) xx 2 - 10 xx 5 xx 2sqrt(2) + 5 xx sqrt(5) xx 4 - 1 xx 4sqrt(2)`

= `25sqrt(5) - 125sqrt(2) + 100sqrt(5) - 100sqrt(2) + 20sqrt(5) - 4sqrt(2)`

= `(25 + 100 + 20)sqrt(5) - (125 + 100 + 4)sqrt(2)`

= `145sqrt(5) - 229sqrt(2)`.

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 1. (ii) | Page 77
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