A) \[{{4}^{2n+1}}\]
B) \[{{4}^{2n}}\]
C) \[{{4}^{2n-1}}\]
D) \[{{4}^{-2n}}\]
Correct Answer: A
Solution :
Since \[(5\sqrt{5}-11)\,(5\sqrt{5}+11)=4\] Þ \[5\sqrt{5}-11=\frac{4}{5\sqrt{5}+11}\], \[\because \,\,\,\,0<5\sqrt{5}-11<1\Rightarrow \,0<{{(5\sqrt{5}-11)}^{2n+1}}<1\], for positive integral n. Again, \[{{(5\sqrt{5}+11)}^{2n+1}}-{{(5\sqrt{5}-11)}^{2n+1}}\] \[=2\left\{ ^{2n+1}{{C}_{1}}{{(5\sqrt{5})}^{2n}}.11+{{\,}^{2n+1}}{{C}_{3}}{{(5\sqrt{5})}^{2n-2}} \right.\] \[\left. \times {{11}^{3}}+....{{+}^{2n+1}}{{C}_{2n+1}}{{11}^{2n+1}} \right\}\] \[=2\,\left\{ ^{2n+1}{{C}_{1}}{{(125)}^{n}}.11{{+}^{2n+1}} \right.{{C}_{3}}{{(125)}^{n-1}}{{11}^{3}}+..\] \[\left. ....{{+}^{2n+1}}{{C}_{2n+1}}{{11}^{2n+1}} \right\}\] = 2k, (for some positive integer k) Let \[{f}'={{(5\sqrt{5}-11)}^{2n+1}}\], then \[[R]+f-{f}'=2k\] Þ \[f-{f}'=2k-[R]\,\,\,\Rightarrow \,\,\,f-{f}'\]is an integer. But, \[0\le f<1;\,0<{f}'<1\,\,\,\Rightarrow -1<f-f'<1\] Þ \[f-{f}'=0\](integer) Þ \[f={f}'\] \[\therefore \,\,Rf=R{f}'={{(5\sqrt{5}+11)}^{2n+1}}{{(5\sqrt{5}-11)}^{2n+1}}\] =\[{{[{{(5\sqrt{5})}^{2}}-{{11}^{2}}]}^{2n+1}}={{4}^{2n+1}}\].
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