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Find the value of (2+5)5+(2-5)5 - Mathematics and Statistics

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Question

Find the value of `(2 + sqrt(5))^5 + (2 - sqrt(5))^5`

Sum

Solution

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

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

Adding (1) and (2), we get,

`(2 + sqrt(5))^5 + (2 - sqrt(5))^5 = 2[""^5"C"_0(2)^5 + ""^5"C"_2(2)^3 (sqrt(5))^2 + ""^5"C"_4(2)(sqrt(5))^4]`

Now 5C0 = 1, 5C4 = 5C1 = 5, 5C2 = `(5 xx 4)/(1 xx 2)` = 10

∴ `(2 + sqrt(5))^5 + (2 - sqrt(5))^5` = 2[1 × 32 + 10 × 8 × 5 + 5 × 2 × 25]

= 2(32 + 400 + 250)

= 2(682)

= 1364

shaalaa.com
Binomial Theorem for Positive Integral Index
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Chapter 4: Methods of Induction and Binomial Theorem - Exercise 4.2 [Page 77]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.2 | Q 3. (ii) | Page 77
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