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The Lucas numbers L(n) part 2

Mathematics
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May 23rd, 2015

L(k)= L(K-1) +L(K-2)

K-1 +  βK-1  + αK-2 +  βK-2 

= αK-2(α +  1) +  βK-2(  β + 1)

= αK—22) + βK-2(  β2)

= αK + βK


May 25th, 2015

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May 23rd, 2015
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May 23rd, 2015
May 25th, 2017
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