17.解:(1)记数列{an}的公差为d,而a3=a1a,-(1分)即(a1+3d)2=(a1+d)(a1+7d),(2分)即a3+6a1d+9d2=a1+8a1d+7d2,故d2-a1d=0,得d=0或d=a1,(3分)因为S20=420,即20a1+190d=420,即2a1+19d=42,4分)当d=0时,a1=21;当d=a1时,a1=2故数列{an}的通项公式为an=21或an=2n,(6分)(2)由(1)可知,an=2n,(7分)1则口a,+1)(a++1)(2n+1)(2n++5-2(2n+1(9分+3故T(3-3)+(5-7)++2n+12n+3-2…十2n+1-2n+3)=2(32n+3)6n+9·(12分)

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