已知数列an满足,a1=1,an>0且a(n+1)×更号下(4+1/an^2)=1(n∈N+)①求数列an的通项公式②数列bn的前n项和Sn满足S(n+1)/an^2=Sn/a(n+1)^2+16n^2-8n-3,若数列bn是等差数列,求b1③在②的条件下,设cn=

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