已知数列{an}的前n和为Sn,且Sn=2an+n^2-3n-2 n为正整数求证:1数列是等比数列2设bn=an*cosn180度,求数
问题描述:
已知数列{an}的前n和为Sn,且Sn=2an+n^2-3n-2 n为正整数求证:1数列是等比数列2设bn=an*cosn180度,求数
列bn的前n项和p
答
s1=a1=2a1+1^2-3*1-2a1=2a1-4a1=4sn=2an+n^2-3n-2s(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2sn-s(n-1)=2an+n^2-3n-2-2a(n-1)-(n-1)^2+3(n-1)+2an=2an-2a(n-1)+n^2-(n-1)^2-3n+3(n-1)an=2an-2a(n-1)+2n-2an=2a(n-1)-2n+2an-2n=...