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Wed 14 Sep, 2005 07:18 pm
Show that if a_n = (1/5^0.5)*(alpha^n - beta^n), where alpha = (1 + 5^0.5) / 2 and beta = (1 - 5^0.5) / 2, then a_n = a_(n-1) + a_(n-2) and a_1 = a_2 = 1. Conclude that f_n = a_n, where f_n is the nth Fibonacci number.
a_n=1/√(5)[ ά^(n-1)+ β^(n-1)]
so
a_n-1+a_n-2=1/√(5)[ ά^(n-1)+ β^(n-1)]+ 1/√(5)[ ά^(n-2)+ β^(n-2)]
a_n-1+a_n-2= 1/√(5)[ ά^(n-2)(ά+1) + β^(n-2)(β+1)]
ά+1=(3+√5)/2=[(1+√5)/2]^2=ά^2
β+1)=(3-√5)/2=[(1-√5)/2]^2= β^2
a_n-1+a_n-2= 1/√(5)[ ά^(n-2)ά^2 + β^(n-2)β^2]= 1/√(5)[ ά^n + β^n]=a_n
&
a_n= a_n-1+a_n-2
Rap c∫;?/
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