（1）当n=1时，a1＝s1＝14a21+12a1−34，解出a1=3，又4Sn=an2+2an-3①当n≥2时4sn-1=an-12+2an-1-3②①-②4an=an2-an-12+2（an-an-1），即an2-an-12-2（an+an-1）=0，∴（an+an-1）（an-an-1-2）=0，∵an+an-1＞0∴a...