设数列{an}的前n项和为Sn,已知a1=1,Sn+1=4an+2,求数列AN的通项公式由已知,a1+a2=4a1+2,故a2=5因Sn+1=4an+2当n>=2时,Sn=4a(n-1)+2两式相减得a(n+1)=4an-4a(n-1),所以a(n+1)-2an=2(an-2an-1)所以{an-2an-1}是以3为首项,2为
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 11:06:02
![设数列{an}的前n项和为Sn,已知a1=1,Sn+1=4an+2,求数列AN的通项公式由已知,a1+a2=4a1+2,故a2=5因Sn+1=4an+2当n>=2时,Sn=4a(n-1)+2两式相减得a(n+1)=4an-4a(n-1),所以a(n+1)-2an=2(an-2an-1)所以{an-2an-1}是以3为首项,2为](/uploads/image/z/634628-20-8.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A5a1%3D1%2CSn%2B1%3D4an%2B2%2C%E6%B1%82%E6%95%B0%E5%88%97AN%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E7%94%B1%E5%B7%B2%E7%9F%A5%2Ca1%2Ba2%3D4a1%2B2%2C%E6%95%85a2%3D5%E5%9B%A0Sn%2B1%3D4an%2B2%E5%BD%93n%3E%3D2%E6%97%B6%2CSn%3D4a%28n-1%29%2B2%E4%B8%A4%E5%BC%8F%E7%9B%B8%E5%87%8F%E5%BE%97a%28n%2B1%29%3D4an-4a%28n-1%29%2C%E6%89%80%E4%BB%A5a%28n%2B1%29-2an%3D2%28an-2an-1%29%E6%89%80%E4%BB%A5%7Ban-2an-1%7D%E6%98%AF%E4%BB%A53%E4%B8%BA%E9%A6%96%E9%A1%B9%2C2%E4%B8%BA)
xUKSY+,5T7?MjgEfT5 Ayd|Qyb+|oI
g2dJ,KGGbv/^Nvu:[uɩNˤocL
h+RT