a1=1,Sn^2-Sn-1^2=an^3,求证数列an为等差数列,并求出通项公式(n>=2,且n属于整数)
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a1=1,Sn^2-Sn-1^2=an^3,求证数列an为等差数列,并求出通项公式(n>=2,且n属于整数)
![a1=1,Sn^2-Sn-1^2=an^3,求证数列an为等差数列,并求出通项公式(n>=2,且n属于整数)](/uploads/image/z/9151952-32-2.jpg?t=a1%3D1%2CSn%5E2-Sn-1%5E2%3Dan%5E3%2C%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97an%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82%E5%87%BA%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%88n%3E%3D2%2C%E4%B8%94n%E5%B1%9E%E4%BA%8E%E6%95%B4%E6%95%B0%EF%BC%89)
Sn^2-Sn-1^2=an^3,
(Sn-Sn-1)(Sn+Sn-1)=an^3,
an(Sn+Sn-1)=an^3,
(Sn+Sn-1)=an²
(Sn-1+Sn-2)=an-1²
两式相减an+an-1=an²-an-1²
an+an-1=(an+an-1)(an-an-1)
an-an-1=1
所以是a1=1,d=1的等差数列
an=a1+(n-1)d=1+n-1=n
(Sn-Sn-1)(Sn+Sn-1)=an^3,
an(Sn+Sn-1)=an^3,
(Sn+Sn-1)=an²
(Sn-1+Sn-2)=an-1²
两式相减an+an-1=an²-an-1²
an+an-1=(an+an-1)(an-an-1)
an-an-1=1
所以是a1=1,d=1的等差数列
an=a1+(n-1)d=1+n-1=n
a1=1,Sn^2-Sn-1^2=an^3,求证数列an为等差数列,并求出通项公式(n>=2,且n属于整数)
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