Chứng minh rằng: 3M < 1/4 - 1/4^10 1/4^1 + 1/4^2 + 1/4^3 + ... + 1/4^10 02/07/2021 Bởi Mackenzie Chứng minh rằng: 3M < 1/4 - 1/4^10 1/4^1 + 1/4^2 + 1/4^3 + ... + 1/4^10
Đáp án: Giải thích các bước giải: $\text{M = $\dfrac{1}{4^1}$ + $\dfrac{1}{4^2}$ + $\dfrac{1}{4^3}$ +…+ $\dfrac{1}{4^{10}}$}$ $\text{4M = $\dfrac{1}{1}$ + $\dfrac{1}{4^1}$ + $\dfrac{1}{4^2}$ +…+ $\dfrac{1}{4^9}$}$ $\text{3M = 4M – M = 1 – $\dfrac{1}{4^{10}}$ = 1 – $\dfrac{1}{4^{10}}$}$ $\text{⇒ M < 1 – $\dfrac{1}{4^{10}}$}$ Bình luận
`M = 1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10` `=> 4M = 4(1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10)` `=> 4M = 1 + 1/4^1 + 1/4^2 + … + 1/4^9` `=> 4M – M = (1 + 1/4^1 + 1/4^2 + … + 1/4^9) + (1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10)` `=> 3M = 1 – 1/4^10``=> M = (1 – 1/4^10)/3` Vì `1 < 4^10` `=> 1 > 1/4^10` `=> 1 – 1/4^10 > 0` `=> 1 – 1/4^10 > (1 – 1/4^10)/3` `=> 1 – 1/4^10 > M` hay `M < 1 – 1/4^10` `=> đpcm` Bình luận
Đáp án:
Giải thích các bước giải:
$\text{M = $\dfrac{1}{4^1}$ + $\dfrac{1}{4^2}$ + $\dfrac{1}{4^3}$ +…+ $\dfrac{1}{4^{10}}$}$
$\text{4M = $\dfrac{1}{1}$ + $\dfrac{1}{4^1}$ + $\dfrac{1}{4^2}$ +…+ $\dfrac{1}{4^9}$}$
$\text{3M = 4M – M = 1 – $\dfrac{1}{4^{10}}$ = 1 – $\dfrac{1}{4^{10}}$}$
$\text{⇒ M < 1 – $\dfrac{1}{4^{10}}$}$
`M = 1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10`
`=> 4M = 4(1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10)`
`=> 4M = 1 + 1/4^1 + 1/4^2 + … + 1/4^9`
`=> 4M – M = (1 + 1/4^1 + 1/4^2 + … + 1/4^9) + (1/4^1 + 1/4^2 + 1/4^3 + … + 1/4^10)`
`=> 3M = 1 – 1/4^10`
`=> M = (1 – 1/4^10)/3`
Vì `1 < 4^10`
`=> 1 > 1/4^10`
`=> 1 – 1/4^10 > 0`
`=> 1 – 1/4^10 > (1 – 1/4^10)/3`
`=> 1 – 1/4^10 > M`
hay `M < 1 – 1/4^10`
`=> đpcm`