so sanh B=1/3^1+1/3^2+1/3^3+…+1/3^99+1/3^100 voi 1/2 30/07/2021 Bởi Margaret so sanh B=1/3^1+1/3^2+1/3^3+…+1/3^99+1/3^100 voi 1/2
Đáp án: Ta có `B = 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^{99} + 1/3^{100} (1)` `=> 3B = 1 + 1/3 + 1/3^2 + …. + 1/3^{98} + 1/3^{99} (2)` Lấy (2) – (1) ta được : `2B = 1 – 1/3^{100}` `=> B = 1/2 – 1/(3^{100}.2) < 1/2` `=> B < 1/2` Giải thích các bước giải: Bình luận
`B=1/3^1 + 1/3^2+1/3^3+….+1/3^99+1/3^100` `=> 3B = 1 + 1/3^1+1/3^2+….+1/3^98+1/3^99` `=> 3B – B = 1 -1/3^100` `=> 2B=1-1/3^100` `=> B = (1-1/3^100)/2= 1/2 -(1/3^100)/2=1/2-2/3^100` Do `2/3^100 > 0 ` nên `B =1/2- 2/3^100` Bình luận
Đáp án:
Ta có
`B = 1/3 + 1/3^2 + 1/3^3 + …. + 1/3^{99} + 1/3^{100} (1)`
`=> 3B = 1 + 1/3 + 1/3^2 + …. + 1/3^{98} + 1/3^{99} (2)`
Lấy (2) – (1) ta được :
`2B = 1 – 1/3^{100}`
`=> B = 1/2 – 1/(3^{100}.2) < 1/2`
`=> B < 1/2`
Giải thích các bước giải:
`B=1/3^1 + 1/3^2+1/3^3+….+1/3^99+1/3^100`
`=> 3B = 1 + 1/3^1+1/3^2+….+1/3^98+1/3^99`
`=> 3B – B = 1 -1/3^100`
`=> 2B=1-1/3^100`
`=> B = (1-1/3^100)/2= 1/2 -(1/3^100)/2=1/2-2/3^100`
Do `2/3^100 > 0 ` nên
`B =1/2- 2/3^100`