數(shù)列{b‹n›}滿足bn=(2n-1)/3ⁿ,求前n項(xiàng)和T‹n›.
T‹n›=(1/3)+(3/3²)+(5/3³)+(7/3⁴)+.+(2n-3)/3ⁿֿ¹+(2n-1)/3ⁿ.(1)
(1/3)T‹n›=(1/3²)+(3/3³)+(5/3⁴)+.+(2n-3)/3ⁿ+(2n-1)/3^(n+1).(2)
(1)-(2)得：（錯(cuò)項(xiàng)相減)
(2/3)T‹n›=(1/3)+2[(1/3²)+(1/3³)+(1/3⁴)+.+(1/3ⁿ)]-(2n-1)/3^(n+1)
=(1/3)+(2/3)[(1/3)+(1/3²)+(1/3³)+(1/3⁴)+.+(1/3ⁿֿ¹)]-(2n-1)/3^(n+1)
=(1/3)+(1/3)[1-(1/3ⁿֿ¹)]-(2n-1)/3^(n+1)
=(2/3)-1/3ⁿ-(2n-1)/3^(n+1)=(2/3)-2(n+1)/3^(n+1)=(2/3)[1-(n+1)/3ⁿ]
∴T‹n›=1-(n+1)/3ⁿ (n=1,2,3,.）