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Question

Let R=(55+11)2n+1 and f=R[R], then prove that Rf=42n+1 .

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Solution

Given R=(55+11)2n+1 where f=R[R]
Let us assume a number f1=(5511)2n+1
5511 lies between 0 and 1
0<5511<1
2n+1 is a positive integer.
0<f1<1
Rf1=(55+11)2n+1(5511)2n+1
2(2n+1C1(55)2n(11)+2n+1C3(55)2n2113+2n+1C5(55)2n3115+...+2n+1C2n+1112n+1)
[R]+ff1=even
ff1=integer
0<f<1 and 0<f1<1
ff1I
ff1=0 f1 cannot be negative or f1 cannot be negative.
f=f1
Now Rf=Rf1 since f=f1
where R=(55+11)2n+1 since Rf=Rf1 where R=(55+11)2n+1
Rf=Rf1=(55+11)2n+1(5511)2n+1
=((55)2121)2n+1 using (a+b)(a+b)=a2b2
=(125121)2n+1=42n+1
Hence proved.


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