A) \[{{2}^{n}}\]
B) \[{{2}^{n+1}}\]
C) \[{{2}^{n-1}}\]
D) \[{{2}^{2n}}\]
Correct Answer: C
Solution :
We know that, \[{{(1+x)}^{n}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+....{{+}^{n}}{{C}_{n}}{{x}^{n}}\] ?.(i) and \[{{(1-x)}^{n}}{{=}^{n}}{{C}_{0}}{{-}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}\] \[+...+{{(-1)}^{n}}{{C}_{n}}{{x}^{n}}\] ?.(ii) On adding Eqs. (i) and (ii), we get \[{{(1+x)}^{n}}+{{(1-x)}^{n}}=2{{[}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{2}}{{x}^{2}}+....{{+}^{n}}{{C}_{n}}{{x}^{n}}]\] \[^{n}{{C}_{0}}{{+}^{n}}{{C}_{2}}+....{{+}^{n}}{{C}_{n}}=\frac{{{2}^{n}}}{2}\] \[={{2}^{n-1}}\]
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