7/22/2019 Quiz A(n)

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Question (Fabio De Girolamo)

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7/22/2019 Quiz A(n)

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Answer (Phung Hoang)

The question can be rewritten as: Give a sequence {A(n)} with initial values A(3)=18, A(4)=32, A(5)=50, A(6)=72,

A(7)=98. Find A(10)

I have posted in my previous comment that A(n)=2 in general, so A(10)=200. This is same as most of your results.

But I think of why the author said about 99.9%. This makes me feel seriously interested in finding more.

The first observation is that the given sequence {A(n)} is not regression, means there is no relation of A(n) and

previous terms. So it is possible to have more than a result. I will prove that, and then put you out of 99.9% failed!

Find a sequence zero Ao(n). I will use the rule of interpolation polynomial Lagrange

 ()  ∑() ∏ − −

=≠

= 

Or in expressed form

 ()  (3) ( − 4)( − 5)( − 6)( − 7)(3 − 4)(3 − 5)(3 − 6)(3 − 7) 

+ (4) ( − 3)( − 5)( − 6)( − 7)(4 − 3)(4 − 5)(4 − 6)(4 − 7) 

+ (5) ( − 3)( − 4)( − 6)( − 7)(5 − 3)(5 − 4)(5 − 6)(5 − 7) 

+ (6) ( − 3)( − 4)( − 5)( − 7)(6 − 3)(6 − 4)(6 − 5)(4 − 7) 

+ (7) ( − 3)( − 4)( − 5)( − 6)(7 − 3)(7 − 4)(7 − 5)(7 − 6) 

Express then shorten, I have ()  2 

Yes, so most of the people have this result A(10) = A0(10) = 2x10x10 = 200. We should be in that 0.1%

But look, I have something different

 () 0() + ( − 3)( − 4)( − 5)( − 6)( − 7) ∗ () where {()} is an any sequence. This can satisfy

the initial condition, while giving us a unlimited number of answers.

As {B(n)} is an any sequence, therefore {A(n)} is also an any sequence.

 ()  2 + ( − 3)( − 4)( − 5)( − 6)( − 7)() 

  Who have A(10) = 200 want to be listed in 0.1%? I select B(n) = constant = 0

  Who have A(10) = 160 want to be listed in 0.1%? I select B(n) = constant = -1/63, or B(n) = (9-n)/63 or any

other that ensure B(10) = -1/63

  Who have A(10) = M want to be listed in 0.1%? You are in if you select B(n) = (M-200)/2520

 ()  2 + − 2002520   ( − 3)( − 4)( − 5)( − 6)( − 7) 

So All of Us are listed in 0.1%. Congratulate!!!

Conclusion: There are unlimited number of sequences if they are non-regression with given initial values.

Thanks,

Phung Hoang