Let n be an integer, and let (Ln) signify the nth Lucas Number.
((Ln)2+(Ln+1)2)2 - 5((L2n+2)*(L2n)-1) = 0
Prove it!
I made use of the page
http://mathworld.wolfram.com/LucasNumber.html
specifically formula [5]
<img src="http://mathworld.wolfram.com/images/equations/LucasNumber/NumberedEquation5.gif" class="numberedequation" alt=" L_n^2-L_(n-1)L_(n+1)=5(-1)^n, " border="0" height="19" width="144">
and formula [21]
<img src="http://mathworld.wolfram.com/images/equations/LucasNumber/NumberedEquation17.gif" class="numberedequation" alt=" L_n^2=L_(2n)+2(-1)^n, " border="0" height="19" width="112">
((L
n)
2+(L
n+1)
2)
2 - 5((L
2n+2)*(L
2n)-1) = 0
(L
2n+2(-1)
n + L
2n+2+2(-1)
n+1)
2 - 5(L
2n+2)*(L
2n) + 5 = 0
(L
2n + L
2n+2)
2 - 5(L
2n+2)*(L
2n) + 5 = 0
(L
2n)
2+2L
2n*L
2n+2+(L
2n+2)
2- 5(L
2n+2)*(L
2n) + 5 = 0
(L
2n)
2-3L
2n*L
2n+2+(L
2n+2)
2 + 5 = 0
L
2n(L
2n-L
2n+2)+L
2n+2(L
2n+2-L
2n) - L
2n*L
2n+2 + 5=0
L
2n(-L
2n+1)+L
2n+2(L
2n+1)- L
2n*L
2n+2 + 5=0
(L
2n+1)(L
2n+2-L
2n)
- L
2n*L
2n+2 + 5=0
(L
2n+1)(L
2n+1)
- L
2n*L
2n+2 + 5=0
(L
2n+1)
2 - L
2n*L
2n+2 + 5=0
[5(-1)
2n+1+L
2n*L
2n+2 ]- L
2n*L
2n+2 + 5=0
-5 + 5 = 0
0=0 QED
|
|
Posted by Jer
on 2013-12-16 09:33:17 |
Long solution.
