A) \[AB=\frac{1}{4}{{(x-a)}^{2n}}-{{(x+a)}^{2n}}\]
B) \[2AB={{(x+a)}^{2n}}-{{(x-a)}^{2n}}\]
C) \[4AB={{(x+a)}^{2n}}-{{(x-a)}^{2n}}\]
D) None of these
Correct Answer: C
Solution :
\[{{(x+a)}^{n}}={{\,}^{n}}{{C}_{0}}{{x}^{n}}+{{}^{n}}{{C}_{1}}{{x}^{n-1}}a+{{\,}^{n}}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+{{\,}^{n}}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+.....\] But by the condition, \[A={{\,}^{n}}{{C}_{0}}{{x}^{n}}+{{\,}^{n}}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+{{\,}^{n}}{{C}_{4}}{{x}^{n-4}}{{a}^{4}}+......\] and \[B={{\,}^{n}}{{C}_{1}}{{x}^{n-1}}a+{{\,}^{n}}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+......\] Hence \[AB=\frac{1}{4}\left\{ {{(x+a)}^{2n}}-{{(x-a)}^{2n}} \right\}\] or \[4AB={{(x+a)}^{2n}}-{{(x-a)}^{2n}}\]
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