Chứng minh rằng: 1/2^2+1/2^3+1/2^4+…+1/2^n<1 help 04/10/2021 Bởi Arya Chứng minh rằng: 1/2^2+1/2^3+1/2^4+…+1/2^n<1 help
Đáp án + giải thích bước giải : Đặt `A = 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^n` `-> 2A = 1/2 + 1/2^2 + 1/2^3 + ….. + 1/2^{n – 1}` `-> 2A – A = (1/2 + 1/2^2 + 1/2^3 + ….. + 1/2^{n – 1}) – (1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^n)` `-> A = 1/2 – 1/2^n` `-> A = 1/2 – 1/2^n < 1/2` `-> 1/2^2 + 1/2^3 + 1/2^4 + .. + 1/2^n < 1 (đpcm)` Bình luận
Đặt A = $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + $\frac{1}{2^4}$ + ….+ $\frac{1}{2^n}$ 2A = $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + …. + $\frac{1}{2^n-^1}$ 2A – A = $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n-^1}$ – ( $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n}$ ) A = $\frac{1}{2}$ – $\frac{1}{2^n}$ Vì $\frac{1}{2}$ – $\frac{1}{2^n}$ < $\frac{1}{2}$ Mà $\frac{1}{2}$ < 1 Nên : $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n}$ < 1 => đpcm Bình luận
Đáp án + giải thích bước giải :
Đặt `A = 1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^n`
`-> 2A = 1/2 + 1/2^2 + 1/2^3 + ….. + 1/2^{n – 1}`
`-> 2A – A = (1/2 + 1/2^2 + 1/2^3 + ….. + 1/2^{n – 1}) – (1/2^2 + 1/2^3 + 1/2^4 + … + 1/2^n)`
`-> A = 1/2 – 1/2^n`
`-> A = 1/2 – 1/2^n < 1/2`
`-> 1/2^2 + 1/2^3 + 1/2^4 + .. + 1/2^n < 1 (đpcm)`
Đặt A = $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + $\frac{1}{2^4}$ + ….+ $\frac{1}{2^n}$
2A = $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + …. + $\frac{1}{2^n-^1}$
2A – A = $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n-^1}$ – ( $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n}$ )
A = $\frac{1}{2}$ – $\frac{1}{2^n}$
Vì $\frac{1}{2}$ – $\frac{1}{2^n}$ < $\frac{1}{2}$
Mà $\frac{1}{2}$ < 1
Nên : $\frac{1}{2}$ + $\frac{1}{2^2}$ + $\frac{1}{2^3}$ + … + $\frac{1}{2^n}$ < 1
=> đpcm