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在数列{an}中,a1=0,且对任意(k∈N*),a2k-1,a2k,a2k+1成等差数列...
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在数列{a
n}中,a
1=0,且对任意(k∈N*),a
2k-1,a
2k,a
2k+1成等差数列,其公差为d
k.
(Ⅰ)若d
k=2k,证明a
2k,a
2k+1,a
2k+2成等比数列(k∈N*);
(Ⅱ)若对任意k∈N*,a
2k-1,a
2k,a2k+1成等比数列,其公比为q
k.
(i)设q
1≠1.证明
是等差数列;
(ii)若a
2=2,证明
(n≥2)
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