数列
已知在正项数列{an} 中,表示前项和且2√sn=an+1,求an
An=2n-1. 4S(n)=(An+1)^2,4S(n-1)=[A(n-1)+1]^2. 4S(n)-4S(n-1)=4An=(An+1)^2-[A(n-1)+1]^2 4An移到等式右边得:(An-1)^2-[A(n-1)+1]^2=0 所以An=-A(n-1)(舍), 或An=A(n-1)+2 再由4S(1)=[A(1)+1]^2=4A(1),解得A(1)=1,所以,An=2n-1
答:因为An=2Sn^2/(Sn-1),而Sn-S(n-1)=An 所以Sn-S(n-1)=2Sn^2/(Sn-1),所以2Sn^2-Sn-2SnS(n-1)+S(...详情>>
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