1.已知：数列an中a1=1,当n≥2时,其前n项和满足sn²=an[sn-(1/2)]；求：an表达式.代入an=sn-s(n-1)到sn²=an[sn-(1/2)],化简得(1/sn)-[1/s(n-1)]=2,而1/s1=1/a1=1,则1/sn是以1为首项,公差为2的等差数列,则1/sn=1+(n-1)×2=2n-1,则sn=1/(2n-1)；验证：a1=s1=1/(2×1)=1成立,则an=sn-s(n-1)=-2/[(2n-1)(2n-3)]；即an=-2/[(2n-1)(2n-3)]
2.bn=1//[(2n-1)(2n+1)]=1/2*2[(2n-1)(2n+1)]=1/2*[(2n+1)-(2n+1)]/[(2n-1)(2n+1)]=1/2*{(2n+1)/[(2n-1)(2n+1)]-(2n+1)/[(2n-1)(2n+1)]}=1/2*[1/[(2n-1)-1/(2n+1)]所以Tn=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/[(2n-1)-1/(2n+1)]=1/2*(1-1/(2n+1)]=n/(2n+1)
3.没有了