The value of ( √ 2 + 1 ) 6 + ( √ 2 − 1 ) 6 will be Options: A . &nbsp -198 B . &nbsp 198 C . &nbsp 99 D . &nbsp -99

Answer: Option B : B ( x + a ) n + ( x − a ) n = 2 [ x n + n C 2 x n − 2 a 2 + n C 4 x n − 4 a 4 + n C 6 x n − 6 a 6 + . . . . . . . ] Here, n = 6, x = √ 2 , a = 1; 6 C 2 = 15, 6 C 4 = 15, 6 C 6 = 1 ∴ ( √ 2 + 1 ) 6 + ( √ 2 − 1 ) 6 = 2 [ ( √ 2 ) 6 + 15. ( √ 2 ) 4 .1 + 15 ( √ 2 ) 2 .1 + 1.1 ] = 2 [ 8 + 15 × 4 + 15 × 2 + 1 ] = 198

The value of (√2+1)6 + (√2−1)6 will be

Discussion Forum : Binomial Theorem

Question -

The value of ${(\sqrt{2} + 1)}^{6}$ + ${(\sqrt{2} - 1)}^{6}$ will be

Options:

A . -198

B . 198

C . 99

D . -99

Answer: Option B
:
B

$(x + a)^{n}$ + $(x - a)^{n}$ = 2 $[x^{n} +^{n} C_{2} x^{n - 2} a^{2} +^{n} C_{4} x^{n - 4} a^{4} +^{n} C_{6} x^{n - 6} a^{6} + . . . . . . .]$

Here, n = 6, x = $\sqrt{2}$ , a = 1;

$^{6} C_{2}$ = 15, $^{6} C_{4}$ = 15, $^{6} C_{6}$ = 1

∴ ( $\sqrt{2} + 1)^{6}$ + ( $\sqrt{2} - 1)^{6}$ = 2 $[(\sqrt{2})^{6} + 15. (\sqrt{2})^{4} .1 + 15 (\sqrt{2})^{2} .1 + 1.1]$

$= 2 [8 + 15 \times 4 + 15 \times 2 + 1] = 198$

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