Tính tổng B=1/3-2/3^2+3/3^3-4/3^4+..+99/3^99-100/3^100 14/08/2021 Bởi Raelynn Tính tổng B=1/3-2/3^2+3/3^3-4/3^4+..+99/3^99-100/3^100
Đáp án+Giải thích các bước giải: `B=1/3-2/3^2+3/3^3-4/3^4+….+9/3^99-100/3^100` `=>3B=1-2/3+3/3^2-4/3^3+……+9/3^98-100/3^99` `=>3B+B=4B=\underbrace{1-1/3+1/3^2-1/3^3+…….+1/3^98-1/3^99}_{A}-100/3^100` Ta có: `3A=3-1+1/3-1/3^2+…..+1/3^97-1/3^98` `=>3A+A=4A=3-1/3^99` `=>A=(3-1/3^99)/4` `=>4B=(3-1/3^99)/4-100/3^100` `=>B=((3-1/3^99)/4-100/3^100)/4` Bình luận
Đáp án+Giải thích các bước giải:
`B=1/3-2/3^2+3/3^3-4/3^4+….+9/3^99-100/3^100`
`=>3B=1-2/3+3/3^2-4/3^3+……+9/3^98-100/3^99`
`=>3B+B=4B=\underbrace{1-1/3+1/3^2-1/3^3+…….+1/3^98-1/3^99}_{A}-100/3^100`
Ta có:
`3A=3-1+1/3-1/3^2+…..+1/3^97-1/3^98`
`=>3A+A=4A=3-1/3^99`
`=>A=(3-1/3^99)/4`
`=>4B=(3-1/3^99)/4-100/3^100`
`=>B=((3-1/3^99)/4-100/3^100)/4`