→ wu1212121212:就是把4^n想辦法湊得跟sin的角度港款 12/27 22:58
推 contentt:有~湊出來了~干溫 12/28 16:53
設 a_0 = 1/2 a_n = [(1+a_n-1)/2]^(1/2) n = 1.2.3...
求 lim 4^n(1-a_n)
n->oo
<sol>
設 a_0 = cosx
=> a_1 = cos(x/2) ， a_2 = cos(x/4) ...
∴ a_0=cos(π/3) a_n=cos(π/3*2^n)
lim (4^n)(1-a_n) = lim (4^n){2sin[π/3*2^(n+1)]}^2 (1)
n->oo n->oo
= lim 2{sin(π/3*2^(n+1)/[π/3*2^(n+1)]}^2 (π/6)^2 (2)
n->oo
= (π)^2/18
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