已知数列{an}满足a1=2,an+1-an=an+1*an,那么a31等于
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已知数列{an}满足a1=2,an+1-an=an+1*an,那么a31等于
![已知数列{an}满足a1=2,an+1-an=an+1*an,那么a31等于](/uploads/image/z/3758566-22-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D2%2Can%2B1-an%3Dan%2B1%2Aan%2C%E9%82%A3%E4%B9%88a31%E7%AD%89%E4%BA%8E)
两边同除an*an+1得:1/an-1/an+1=1
1/an+1-1/an=-1,所以数列{1/an}为等差数列
1/an=1/a1+(-1)*(n-1)
1/a31=1/2+(-1)*30
1/a31=-59/2
a31=-2/59
1/an+1-1/an=-1,所以数列{1/an}为等差数列
1/an=1/a1+(-1)*(n-1)
1/a31=1/2+(-1)*30
1/a31=-59/2
a31=-2/59
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