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Wednesday, March 30, 2016

2016/030) Let R= (5\sqrt{5} + 11)^{2n+1} and f= R- \lfloor R \rfloor where \lfloor R \rfloor is greatest integer functon. Prove that Rf = 4^{2n+1}

we have (5\sqrt{5} - 11) = \frac{(5\sqrt{5} - 11)(5\sqrt{5} - 11)}{5\sqrt{5} + 11}
= \frac{125-121}{5\sqrt{5} + 11} = = \frac{4}{5\sqrt{5} + 11} < 1
R-r =   (5\sqrt{5} + 11)^{2n+1} -  (5\sqrt{5} - 11)^{2n+1}
= \sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i 11^{2n+1-i} -\sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i  (-11)^{2n+1-i}
= 2\sum \limits_{i=0}^n{2n+1\choose i} (5\sqrt{5})^{2i} 11^{2n+1-1}
= 2\sum \limits_{i=0}^n{2n+1\choose i} 125^i 11^{2n+1-1} an integer
so r = R -\lfloor R \rfloor
So r = f = (5\sqrt{5} - 11)^{2n+1}
Hence
Rf = (5\sqrt{5} +  11)^{2n+1}(5\sqrt{5} - 11)^{2n+1}
= ((5\sqrt{5} + 11)(5\sqrt{5} - 11))^{2n+1} = (125-121)^{2n+1} = 4^{2n+1}

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