pure maths ------ binomial

( √3 ± √2 )2n = (5 ± 2√6)n
Note that ( an ± bn√c ) n
= (an2 + bn2c ± anbn√c) n-1
= ( an-1 ± bn-1√c ) n-1 where an-1=an2 + bn2c, bn-1=anbn
= ...
= a1 ± b1√c

Hence
( √3 + √2 )2n = pn + qn√6 ............ (1)
( √3 - √2 )2n = pn - qn√6 ............ (2)
(1)+(2),
( √3 + √2 )2n + ( √3 - √2 )2n = 2pn
Since ( √3 - √2 )2n > 0, ( √3 + √2 )2n < 2pn
Since ( √3 - √2 )2n < 1, ( √3 + √2 )2n > 2pn - 1
Hence the proof.

2007-10-16 23:56:57 補充：
Line 3 should be= (a[n]^2 b[n]^2 c ± a[n]b[n]√c) ( a[n] ± b[n]√c ) ^(n-2)Line 4 should be= ( a[n-1] ± b[n-1]√c ) ( a[n] ± b[n]√c ) ^(n-2)