Determine the value of ( 2 1 + 5 ) 1 0 + ( 2 1 − 5 ) 1 0 .
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Sir, I am studying 12th grade under Tamil nadu state board syllabus, To say frankly I couldn't understand the solution for this problem, so please give me a simpler or a brief solution to understand the problem easier (if u can).
brilliant solution. but did you know by default the roots of the quadratic equation and worked from there ?
Note that φ = 2 1 + 5 is the golden ratio, 2 1 − 5 = − φ 1 , and φ + ( − φ 1 ) = 1 . Then we have
( φ − φ 1 ) 2 φ 2 − 2 + φ 2 1 ⟹ φ 2 + φ 2 1 ( φ 2 + φ 2 1 ) ( φ − φ 1 ) φ 3 − φ + φ 1 − φ 2 1 ⟹ φ 3 − φ 3 1 ( φ 3 − φ 3 1 ) ( φ 2 + φ 2 1 ) φ 5 + φ − φ 1 − φ 5 1 ⟹ φ 5 − φ 5 1 ( φ 5 − φ 5 1 ) 2 φ 1 0 − 2 + φ 1 0 1 ⟹ φ 1 0 + ( − φ 1 ) 1 0 = 1 2 = 1 = 3 = 3 × 1 = 3 = 4 = 4 × 3 = 1 2 = 1 1 = 1 1 2 = 1 2 1 = 1 2 3
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We'll generalize it , think of the quadratic equation whose roots are
α = 2 1 + 5 and β = 2 1 − 5
The quadratic equation will be x 2 − x − 1 = 0
Now, make recurrence relation from this quadratic equation, such that a n = α n + β n
This is possible, see that if you make the quadratic to become the characteristic equation, then recurrence will be a n − a n − 1 − a n − 2 = 0 , or more simple looking, a n = a n − 1 + a n − 2
And, now , by subsituting n = 0 and n = 1 , for general form, we get
a 0 = α 0 + β 0 = 1 + 1 = 2
a 1 = α + β = 1
Now, from the recurrence, we conclude that add adjacent two terms to get the next get third term, add 2nd and 3rd to get 4th, and so on, so the sequence is 2 , 1 , 3 , 4 , 7 , 1 1 , 1 8 , 2 9 , 4 7 , 7 6 , 1 2 3 , giving 1 0 t h term as 1 2 3
This sequence is actually Luca's sequence.