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Evaluate (√2 + 1)^5 − (√2 − 1)^5 using binomial.

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Evaluate (√2 + 1)5 − (√2 − 1)5 using binomial.

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(x+ 1)5 =x5 + 5C1 x4 .1 + 5C2 x3 .12 + 5C3 x2 .13 + 5C4x1 .14 + 5C5x0 .15 (i) 

= x5 + 5x4 + 10x3 + 10x2 + 5x + 1 (x– 1)5 = x5 + 5C1x4 (– 1) + 5C2x3 (– 1)2 + 5C3 x2 (– 1)3 + 5C4x(– 1)4 + 5C5x0 (– 1)5 (ii) 

= x5 - 5x4 + 10x3 – 10x2 + 5x – 1 

Subtracting (ii) from (i), 

we get (x + 1)5 – (x– 1)5 = 2[5x4 + 10x2 + 1] 

Put x= √2, we get

(√2 + 1)5 − (√2 − 1)5

2 [5 (√2) 4 + 10 (√2) 2 + 1] 

= 2[20 + 20 + 1]

= 2 × 41 = 82

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