Maharashtra State BoardHSC Science (General) 11th
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Prove that (5+1)5-(5-1)5 = 352 - Mathematics and Statistics

Sum

Prove that `(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5` = 352

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Solution

`(sqrt(5) + 1)^5 = ""^5"C"_0 (sqrt(5))^5 + ""^5"C"_1 (sqrt(5))^4 (1) + ""^5"C"_2 (sqrt(5))^3 (1)^2  + ""^5"C"_3 (sqrt(5))^2 (1)^3 + ""^5"C"_4 (sqrt(5)) (1)^4 + ""^5"C"_3 (1)^5`    ...(1)

and `(sqrt(5) + 1)^5 = ""^5"C"_0(sqrt(5))^5 - ""^5"C"_1(sqrt(5))^4(1) + ""^5"C"_2 (sqrt(5))^3(1)^2 - ""^5"C"_3 (sqrt(5))^2(1)^3 + ""^5"C"_4 (sqrt(5))(1)^4 - ""^5"C"_5 (1)^5`  ...(2)

Subtracting (2) from (1), we get,

`(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5 = 2[""^5"C"_1 (sqrt(5))^4 (1) + ""^5"C"_3 (sqrt(5))^2 (1)^3 + ""^5"C"_5(1)^5]`

Now, 5C1 = 5

5C3 = `(5 xx 4 xx 3)/(1 xx 2 xx 3)` = 10

5C5 = 1

∴ `(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5` = 2[5 × 25 × 1 + 10 × 5 × 1 + 1 × 1]

= 2[125 + 50 + 1]

= 2(176)

= 352

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 4. (ii) | Page 77
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