设数列｛an｝的前n项和Sn=2n2,｛bn｝为等比数列,且a1=b1,b2（a2-a1）=b1,求｛an｝｛bn｝通项公式

先求{an} an=Sn-S(n-1)
=2n²-2(n-1)²
=2n²-2n²+4n-2
=4n-2
有{an}, an=4n-2
b1=a1=4×1-2=S1=2×1²=2
a2=4×2-2=S2-a1=2×2²-2=6
b2×(6-2)=b1
→b2=b1/4
{bn}为等比数列
→bn=b1/4^(n-1)
=2/2^(2n-2)
=2^(3-2n)
得{bn}bn=2^(3-2n)