A) \[{{2}^{4n-2}}+{{(-1)}^{n}}{{2}^{2n-1}}\]
B) \[{{2}^{4n-2}}+{{2}^{2n-1}}\]
C) \[{{2}^{2n-1}}+{{(-1)}^{n}}\,{{2}^{4n-2}}\]
D) None of these
Correct Answer: A
Solution :
We have \[^{4n}{{C}_{0}}+{{\,}^{4n}}{{C}_{2}}{{x}^{2}}+{{\,}^{4n}}{{C}_{4}}{{x}^{4}}+...+{{\,}^{4n}}{{C}_{4n}}{{x}^{4n}}\]\[=\frac{1}{2}[{{(1+x)}^{4n}}+{{(1-x)}^{4n}}]\] Putting \[x=1\] and x = i, we get \[^{4n}{{C}_{0}}+{{\,}^{4n}}{{C}_{2}}+{{\,}^{4n}}{{C}_{4}}+...+{{\,}^{4n}}{{C}_{4n}}=\frac{1}{2}[{{2}^{4n}}]\] and \[^{4n}{{C}_{0}}-{{\,}^{4n}}{{C}_{2}}+{{\,}^{4n}}{{C}_{4}}-...+{{\,}^{4n}}{{C}_{4n}}\] =\[\frac{1}{2}[{{(1+i)}^{4n}}+{{(1-i)}^{4n}}]\] Thus, \[2{{[}^{4n}}{{C}_{0}}+{{\,}^{4n}}{{C}_{4}}+...+{{\,}^{4n}}{{C}_{4n}}]\]\[={{2}^{4n-1}}+\frac{1}{2}{{[{{(1+i)}^{4n}}+(1-i)]}^{4n}}\] Now, \[{{(1+i)}^{4n}}+{{(1-i)}^{4n}}={{\left[ \sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right) \right]}^{4n}}\]\[+{{\left[ \sqrt{2}\left( \cos \frac{\pi }{4}-i\sin \frac{\pi }{4} \right) \right]}^{4n}}\] \[={{2}^{2n}}(\cos n\pi +i\sin n\pi )+{{2}^{2n}}(\cos n\pi -i\sin n\pi )\] \[={{2}^{2n+1}}\cos n\pi ={{2}^{2n+1}}{{(-1)}^{n}}\] \[\therefore \]\[2{{[}^{4n}}{{C}_{0}}+{{\,}^{4n}}{{C}_{4}}+...+{{\,}^{4n}}{{C}_{4n}}]={{2}^{4n-1}}+\frac{1}{2}{{2}^{2n+1}}{{(-1)}^{n}}\] Þ \[^{4n}{{C}_{0}}+{{\,}^{4n}}{{C}_{4}}+...+{{\,}^{4n}}{{C}_{4n}}={{2}^{4n-2}}+{{(-1)}^{n}}{{2}^{2n-1}}\] Trick: Check by putting n = 1, 2.
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