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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find $(x+1)^6+(x-1)^6$ Hence or otherwise evaluate $(\sqrt 2+1)^6+(\sqrt 2-1)^6$

$\begin{array}{1 1}(A)\;190\\(B)\;195\\(C)\;196\\(D)\;198\end{array} $

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1 Answer

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Toolbox:
  • $(1+x)^n=nC_0+nC_1x+nC_2x^2+....nC_1X^r+....+nC_nx^n$
$(x+1)^6=x^6+6C_1x^5.1+6C_2.x^4.1^2+6C_3.x^3.1^3+6C_4.x^2.1^4+6C_5.x.1^5+6C_6.x^4.1^6$
$\Rightarrow x^6+6x^5+15x^4+20x^3+15x^2+6x+1$-----(1)
$(x-1)^6=x^6+6C_1x^5.(-1)+6C_2.x^4.(-1)^2+6C_3.x^3.(-1)^3+6C_4.x^2.(-1)^4+6C_5.x.(-1)^5+6C_6.x^0.(-1)^6$
$\Rightarrow x^6-6x^5+15x^4-20x^3+15x^2-6x+1$-----(2)
Adding (1) and (2)
$(x+1)^6+(x-1)^6=2[x^6+15x^4+15x^2+1]$
Putting $x=\sqrt 2$
$(\sqrt 2+1)^6+(\sqrt 2-1)^6=2[(\sqrt 2)^6+15(\sqrt 2)^4+15(\sqrt 2)^2+1]$
$\Rightarrow 2[8+60+30+1]$
$\Rightarrow 2[99]$
$\Rightarrow 198$
Hence (D) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 
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