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Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate (2+1)6+(2-1)6

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Question

Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate `(sqrt2 + 1)^6 + (sqrt2 -1)^6`

Sum
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Solution

Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as

`(x + 1)^6  = x^6  +  ^6C_1  x^5  1^1 +  ^6C_2  x^4  xx 1^2  + ^6C_3  x^3  xx  1^3  +  ^6C_4  x^2  1^4  +  ^6C_5. x. 1^5  + 1^6`

= `x^6  + 6x^5  + 15x^4  + 20x^3  + 15x^2  + 6x + 1`

similarly, `(x -1)^6  = x^6  - 6x^5  + 15x^4  - 20x^3  + 15x^2 - 6x + 1`

on adding `(x - 1)^6  + (x - 1)^6 = 2(x^6 + 15x^4 + 15x^2 + 1)`

Putting x = `sqrt2` in this

`(sqrt2 + 1)^6  + (sqrt2 - 1)^6 = 2[(sqrt2)^6  + 15 (sqrt2)^4  + 15 (sqrt2)^2 + 1]`

= 2[8 + 15 x 4 + 15 x 2 +1]

= 2[8 + 60 + 30 +1]

= 2 x 99

= 198

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Chapter 7: Binomial Theorem - EXERCISE 7.1 [Page 133]

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NCERT Mathematics [English] Class 11
Chapter 7 Binomial Theorem
EXERCISE 7.1 | Q 12. | Page 133

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