灰儿0919
幼苗
共回答了32个问题采纳率:81.3% 举报
an=3a(n-1)-2n+3=3a(n-1)-3n+n+3=3a(n-1)-3(n-1)+n
an-n=3a(n-1)-3(n-1)
[an-n]/[a(n-1)-(n-1)]=3,为定值,又a1=-1,不等于0,数列有意义.
a1-1=-1-1=-2
{an-n}是等比数列,首项为-2,公比为3.
a1+a2+...+an
=(a1-1)+(a2-2)+...+(an-n)+(1+2+...+n)
=(-2)(3^n-1)/(3-1)+n(n+1)/2
=1-3^n+n(n+1)/2
1年前
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