求n+1/2*3^n-1的前n项和

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求n+1/2*3^n-1的前n项和

(n+1)/(2*3^n-1) 是不是 (n+1)/(2*3^(n-1)) n-1是3的n-1次方吧
再问：是啊= 0=
再答：设Sn =(1+1)/(2*3^(0))+(2+1)/(2*3)+......+(n+1)/(2*3^(n-1)) 则3Sn=3*[(1+1)/(2*3^(0))+(2+1)/(2*3)+......+(n+1)/(2*3^(n-1))] 3Sn-Sn=3*(1+1)/(2*3^0) +[3*(2+1)/(2*3)-(1+1)/(2*3^0)]+[3*(3+1)/2*3^2)-(2+1)/(2*3)]+.... +[3*(n+1)/(2*3^(n-1))-(n-1+1)/(2*3^(n-2))]-(n+1)/(2*3^(n-1)) =3+1/2 +1/(2*3)+1/(2*3^2)+.....+1/(2*3^(n-2)) -(n+1)/(2*3^(n-1)) =3+1/2(1+1/3+1/3^2+...+1/3^(n-2) ]-(n+1)/(2*3^(n-1)) =3+1/2 *(1-1/3^(n-1))/(1-1/3) -(n+1)/(2*3^(n-1)) =3+1/2 *3/2 *(1-1/3^(n-1)) -(n+1)/(2*3^(n-1)) =3+3/4 *(1-1/3^(n-1)) -(n+1)/(2*3^(n-1)) =3+1/(2*3^(n-1)) *[-3/2-n-1] +3/4 =15/4 -(n+5/2)/(2*3^(n-1))=2Sn 所以Sn=15/8 -(n+5/2)/(4*3^(n-1))

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