Sequence : S_{n+3} = 2(S_{n+2}) + 2(S_{n+1}) – S_{n}

 
 
The sequence   F_{n}   of Fibonacci numbers is defined by the recurrence relation

F_{n} = F_{n-1} + F_{n-2}

with seed values
F_1 = F_2 = 1,    or
F_0 = 0,    F_1 = 1

 

Let   S_1 = S_2 = 1,    S_3 = 4,   and

S_{n+3} = 2 \cdot S_{n+2} + 2 \cdot S_{n+1} - S_{n}     for   n \geq 1

 

S_1 = S_2 = 1^2 = ( \,F_1 \,)^2 = ( \,F_2 \,)^2     S_3 = 2^2 = ( \,F_3 \,)^2

S_4 = 2 \cdot S_{3} + 2 \cdot S_{2} - S_{1} = 2(2^2) + 2(1^2) - 1^2 = 3^2 = ( \,F_4 \,)^2
S_5 = 2 \cdot S_{4} + 2 \cdot S_{3} - S_{2} = 2(3^2) + 2(2^2) - 1^2 = 5^2 = ( \,F_5 \,)^2
S_6 = 2 \cdot S_{5} + 2 \cdot S_{4} - S_{3} = 2(5^2) + 2(3^2) - 2^2 = 8^2 = ( \,F_6 \,)^2
S_7 = 2 \cdot S_{6} + 2 \cdot S_{5} - S_{4} = 2(8^2) + 2(5^2) - 3^2 = 13^2 = ( \,F_7 \,)^2
S_8 = 2 \cdot S_{7} + 2 \cdot S_{6} - S_{5} = 2(13^2) + 2(8^2) - 5^2 = 21^2 = ( \,F_8 \,)^2
S_9 = 2 \cdot S_{8} + 2 \cdot S_{7} - S_{6} = 2(21^2) + 2(13^2) - 8^2 = 34^2 = ( \,F_9 \,)^2

Is   S_{p}   always a perfect square?   Yes.

Suppose this is true for   p \; \leq \; (n+2).
Let’s prove that this result also holds for   p \; = \; n+3

F_{n+2} \; = \; F_{n+1} \; + \; F_{n}

S_{n+3}
= 2 \cdot S_{n+2} + 2 \cdot S_{n+1} - S_{n}
= 2( \,F_{n+2} \,)^2 + 2( \,F_{n+1} \,)^2 - ( \,F_{n} \,)^2
= 2( \,F_{n+2} \,)^2 + 2( \,F_{n+1} \,)^2 - ( \,F_{n+2} - F_{n+1} \,)^2
= 2( \,F_{n+2} \,)^2 + 2( \,F_{n+1} \,)^2 - ( \,F_{n+2} \,)^2 - ( \,F_{n+1} \,)^2 + 2 \cdot F_{n+2} \cdot F_{n+1}
= ( \,F_{n+2} \,)^2 + ( \,F_{n+1} \,)^2 + 2 \cdot F_{n+2} \cdot F_{n+1}
= ( \,F_{n+2} + F_{n+1} \,)^2
= ( \,F_{n+3} \,)^2

 

S_{p}   is a perfect square for all   p.
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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