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Question: Let \(R = (5\sqrt{5} + 11)^{2n + 1}\) and \(f = R - \lbrack R\rbrack\) where [.] denotes the greates...

Let R=(55+11)2n+1R = (5\sqrt{5} + 11)^{2n + 1} and f=R−[R]f = R - \lbrack R\rbrack where [.] denotes the greatest integer function. The value of R.f is

A

42n+14^{2n + 1}

B

42n4^{2n}

C

42n−14^{2n - 1}

D

4−2n4^{- 2n}

Answer

42n+14^{2n + 1}

Explanation

Solution

Since f=R−[R]f = R - \lbrack R\rbrack, R=f+[R]R = f + \lbrack R\rbrack

[55+11]2n+1=f+[R]\lbrack 5\sqrt{5} + 11\rbrack^{2n + 1} = f + \lbrack R\rbrack, where [R] is integer

Now let f′=[55−11]2n+1,0<f′<1f' = \lbrack 5\sqrt{5} - 11\rbrack^{2n + 1},0 < f' < 1

f+[R]−f′=[55+11]2n+1−[55−11]2n+1f + \lbrack R\rbrack - f^{'} = \lbrack 5\sqrt{5} + 11\rbrack^{2n + 1} - \lbrack 5\sqrt{5} - 11\rbrack^{2n + 1}=2[2n+1C1(55)2n(11)1+2n+1⥂C3(55)2n−2(11)3+.......]2\left\lbrack 2n + 1C_{1}(5\sqrt{5})^{2n}(11)^{1} +^{2n + 1} ⥂ C_{3}(5\sqrt{5})^{2n - 2}(11)_{}^{3} + ....... \right\rbrack

=2.(Integer)2.(\text{Integer}) = 2K(K∈N)2K(K \in N) = Even integer

Hence f−f′f - f' = even integer ­– [R], but −1<f−f′<1- 1 < f - f' < 1. Therefore, f−f′=0f - f' = 0 ∴f=f′\therefore f = f'

Hence R.f = R.f1=(55+1)2n+1(55−11)2n+1=42n+1R.f^{1} = (5\sqrt{5} + 1)^{2n + 1}(5\sqrt{5} - 11)^{2n + 1} = 4^{2n + 1}.