A=1- $\frac{3}{4}$ + $(\frac{3}{4})^{2}$ – $(\frac{3}{4})^{3}$+ $(\frac{3}{4})^{4}$ – … – $(\frac{3}{4})^{2009}$ + $(\frac{3}{4})^{2010}$ Giải đún

Bởi

A=1- $\frac{3}{4}$ + $(\frac{3}{4})^{2}$ – $(\frac{3}{4})^{3}$+ $(\frac{3}{4})^{4}$ – … – $(\frac{3}{4})^{2009}$ + $(\frac{3}{4})^{2010}$
Giải đúng nhất hợp lý nhất sẽ đc 5 sao ! ^_^

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  1. ` A = 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^{2009} + (3/4)^{2010} `

    ` => 3/4 A = 3/4 – (3/4)^2 + (3/4)^3 – (3/4)^4 + (3/4)^5 – … – (3/4)^{2010} + (3/4)^{2011} `

    ` => A + 3/4 A = 1 – 3/4 + (3/4)^2 – (3/4)^3 + (3/4)^4 – … – (3/4)^{2009} + (3/4)^{2010} `

    ` => 7/4 A = 1 + 3/4 – 3/4 + (3/4)^2 – (3/4)^2 + … + (3/4)^{2010} – (3/4)^{2010} + (3/4)^{2011} `

    ` => 7/4 A = 1 + (3/4)^{2011} `

    ` => A = 4/7 + 4/7 . 3/4 . (3/4)^{2010} `

    ` => A = 4/7 + 3/7 . (3/4)^{2010} `

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  2. Tham khảo

     `A=1-\frac{3}{4}+(\frac{3}{4})^2-….-(\frac{3}{4})^{2009}+(\frac{3}{4})^{2010}`

    `⇒\frac{3}{4}A=\frac{3}{4}-(\frac{3}{4})^2-(\frac{3}{4})^3-…-(\frac{3}{4})^{2010}+(\frac{3}{4})^{2011}`

    `⇒\frac{3}{4}A+A=\frac{3}{4}-(\frac{3}{4})^2-(\frac{3}{4})^3-…-(\frac{3}{4})^{2010}+(\frac{3}{4})^{2011}+(1-\frac{3}{4}+(\frac{3}{4})^2-….-(\frac{3}{4})^{2009}+(\frac{3}{4})^{2010})`

    `⇒\frac{7}{4}A=(\frac{3}{4})^{2011}+1`

    $⇒A=\dfrac{(\dfrac{3}{4})^{2011}+1}{\dfrac{7}{4}}$

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