取倒数,可得1/an=3+1/a(n-1),
即1/an为等差数列,
1/an=3n-2,
an=1/(3n-2)
bn=[1/(3n-2)]*[1/(3n+1)]
=(1/3)*{[1/(3n-2)]-[1/(3n+1)]}
求和可得,sn=n/(3n+1)
取倒数,可得1/an=3+1/a(n-1),
即1/an为等差数列,
1/an=3n-2,
an=1/(3n-2)
bn=[1/(3n-2)]*[1/(3n+1)]
=(1/3)*{[1/(3n-2)]-[1/(3n+1)]}
求和可得,sn=n/(3n+1)