数列an ,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an
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数列an,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an数列an,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求
数列an ,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an
数列an ,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an
数列an ,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an
an=(根号sn+根号sn-1)/2得,2(Sn-Sn-1)=根号sn+根号sn-1
得,根号sn-根号sn-1=1/2
所以 根号sn是等差数列
根号sn=1+(n-1)/2=(n+1)/2 ,Sn=(n+1)^2/4
an=Sn-Sn-1=(2n+1)/4 (n>1)
a1=1
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