Crude solution
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送交者: 青霉竹驴 于 2009-01-07, 07:35:19:
回答: A math question - solution? 由 t1301520 于 2009-01-07, 06:38:43:
Let p_1=p, q_1=q.
Write
p_n=p_{n-1}+rq_{n-1} (1)
and
q_n=q_{n-1}+p_{n-1} (2).
Your problem is to show p_n/q_n = sqrt(r) as n tends to infinity.
Subtracting (1) from (2) yields
p_n - q_n = (r-1)q_{n-1}
which gives
p_n/q_n= 1+(r-1)q_{n-1}/q_n (3).
From (2) it follows
q_n/q_{n-1}=1+p_{n-1}/p_{n-1} (4).
Put s_n = p_n/q_n.
Substituting (4) into (3) and tidying up give
s_n = 1 + (r-1)/(1+s_{n-1}) (5).
Let n tend to infinity in (5), we have
s=1+(r-1)/(1+s) (6)
where s denotes the limit of p_n/q_n.
It's easy to obtain that
s=sqrt(r)
by solving (6).
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