您的举报已经提交成功，我们将尽快处理，谢谢！ 大家还关注 (window.slotbydup=window.slotbydup || []).push({ id: '2081942', container: s, size: '1000,60', display: 'inlay-fix'(window.slotbydup=window.slotbydup || []).push({ id: '2081942', container: s, size: '1000,60', display: 'inlay-fix'数列求和 an=（2n+1）/[2*3^（n-1）]_百度作业帮 数列求和 an=（2n+1）/[2*3^（n-1）] 数列求和 an=（2n+1）/[2*3^（n-1）] 分析：错位相减设前n项和为Sn,则Sn=3/[2*3^0]+5/[2*3^1]+...+(2n+1)/[2*3^(n-1)]∴Sn/3=3/[2*3^1]+5/[2*3^2]+...+(2n+1)/[2*3^n]∴Sn-Sn/3=3/[2*3^0]+2/[2*3^1]+...+2/[2*3^(n-1)]-(2n+1)/[2*3^n]=2/3-(2n+1)/[2*3^n]+[1/(3^1)+1/(3^2)+...+1/(3^(n-1))]=2/3-(2n+1)/[2*3^n]+1/3[1-(1/3)^(n-1)]/[1-1/3]∴可以解得：Sn=3-(n+2)/[2*3^(n-1)] an=（2n+1）/[2*3^（n-1）] = n *(1/3)^(n-1) + (1/2) (1/3)^(n-1)summation {an}=[ summation{n *(1/3)^(n-1)} ] + (3/4) ( 1-(1/3)^n)consider1+x+x^2+..+x^n= (x^(n+1) -1)/(x-1)