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One harder sequence (Posted on 2011-05-24) Difficulty: 4 of 5
Given that n is a whole number, what is the next expression in this series of expressions, each of which is valid for all n?


No Solution Yet Submitted by broll    
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re(3): One possible answer to what's next Comment 6 of 6 |
(In reply to re(2): One possible answer to what's next by Dej Mar)


Let (k)(a)(a-1)=(k+1)(b)(b-1); then (k+1)*(2b-1)^2-k*(2a-1)^2=1.

The first solution is always {a,b}={1,1}, and the second {a,b}={2k+2},{2k+1}. Ignore the solutions {a} for the purpose of clarity.

Denote the function for each row of solutions for successive k as Fr, then for all k the third solution {b} is given by Fr(3{b})=8k^2+6k+1, the fourth Fr(4{b})=32k^3+40k^2+12k+1, etc. as per the series of expressions set out in the problem.

Call the function of the Pth row, Fr(P); then Fr(P) solves for any and all k as the Pth solution of the corresponding Pellian.

Edited on June 27, 2011, 10:25 pm
  Posted by broll on 2011-06-27 22:22:10

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