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If the sum of odd numbered terms and even numbered terms in the expansion of $(x+a)^n$ are $A,B$, then $(x^2-a^2)^n=?$

$\begin{array}{1 1} A^2 - B^2 \\ A^2+B^2 \\ 4AB \\ (A+B)^2 \end{array} $

1 Answer

$(x+a)^n=^nC_0x^n+^nC_1x^{n-1}a+..............$
$=(^nC_0x^n+^nC_2x^{n-2}a^2+..........)+(^nC_1x^{n-1}a+^nC_3x^{n-3}a^3+.....)$
Given : $A=^nC_0x^n+^nC_2x^{n-2}a^2+...............$
$B=^nC_1x^{n-1}.a+^nC_3x^{n-3}a^3+............$
$\Rightarrow\:(x+a)^n=A+B$
$(x-a)^n=^nC_0x^n-^nC_1x^{n-1}a+^nC_2x^{n-2}a^2-.............$
$=(^nC_0x^n+^nC_2x^{n-2}a^2+.........)-(^nC_1x^{n-1}a+^nC_3x^{n-3}a^3+....)$
$\Rightarrow\:(x-a)^n=A-B$
$\therefore\:(x^2-a^2)^n=(x+a)^n.(x-a)^n$
$=(A+B)(A-B)=A^2-B^2$
answered Sep 24, 2013 by rvidyagovindarajan_1
 

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