正项数列an的前n项和sn,满足sn²-(n²+n-1)sn-(n²+n)=0(1)求an
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/20 04:18:29
![正项数列an的前n项和sn,满足sn²-(n²+n-1)sn-(n²+n)=0(1)求an](/uploads/image/z/677120-32-0.jpg?t=%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%2C%E6%BB%A1%E8%B6%B3sn%26%23178%3B-%EF%BC%88n%26%23178%3B%2Bn-1%EF%BC%89sn-%EF%BC%88n%26%23178%3B%2Bn%EF%BC%89%3D0%EF%BC%881%EF%BC%89%E6%B1%82an)
x){v˅;MczbY-O;{"O'<۽Ŷyjʆֺt@yt硈El
glim~
]g$
5lM[ӎ@-@ͽ }`6L @
=y:>lx{)H\HPA`A8g~O7HE$*22yv
正项数列an的前n项和sn,满足sn²-(n²+n-1)sn-(n²+n)=0(1)求an
正项数列an的前n项和sn,满足sn²-(n²+n-1)sn-(n²+n)=0(1)求an
正项数列an的前n项和sn,满足sn²-(n²+n-1)sn-(n²+n)=0(1)求an
Sn^2-(n^2+n+1)Sn-(n^2+n)=0
则(Sn-1)(Sn-(n^2+n))=0
而Sn≠1所以Sn=n^2+n
所以an=Sn-S(n-1)=n^2+n-(n-1)^2-(n-1)=2n