已知数列{an}的前n项和为Sn,且Sn=1/2n^2+11/2n,数列{bn}满足b(n+2)-2b(n+1)+bn=0n∈N*,且b3=11,b1+b2+……+b9=153.求数列{bn}的通项公式.
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/16 12:46:44
![已知数列{an}的前n项和为Sn,且Sn=1/2n^2+11/2n,数列{bn}满足b(n+2)-2b(n+1)+bn=0n∈N*,且b3=11,b1+b2+……+b9=153.求数列{bn}的通项公式.](/uploads/image/z/2117869-61-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%3D1%2F2n%5E2%2B11%2F2n%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3b%28n%2B2%29-2b%28n%2B1%29%2Bbn%3D0n%E2%88%88N%2A%2C%E4%B8%94b3%3D11%2Cb1%2Bb2%2B%E2%80%A6%E2%80%A6%2Bb9%3D153.%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.)
xQJ@~wӸ"y/YȒ("
)((b/҃ÔnҾъIaa替o۸H>NnTyA,h6yJ&)siKz y8z\[ bee ɴwH0hW;Iȼ
4ޕz9lrIEd ĮaƳq<ڜq?D
^|}7ӹmHhn7Ty&uj<ך|Eai c