C = 1/3 + 2/3^2 + 3/3^3 +…+2022/3^2022 So sánh A với 3/4

1 bình luận về “C = 1/3 + 2/3^2 + 3/3^3 +…+2022/3^2022 So sánh A với 3/4”

1. Giải đáp:

    

   Lời giải và giải thích chi tiết:

    C = $\frac{1}{3}$ + $\frac{2}{3^2}$ + $\frac{3}{3^3}$ + … + $\frac{2022}{3^{2022}}$

   3C=1+$\frac{2}{3}$ + $\frac{3}{3^2}$+ … + $\frac{2022}{3^{2021}}$  

   3C-C=( 1+$\frac{2}{3}$ + $\frac{3}{3^2}$+…+$\frac{2022}{3^{2021}}$)-($\frac{1}{3}$ + $\frac{2}{3^2}$ + $\frac{3}{3^3}$ +…+$\frac{2022}{3^{2022}}$)

   2C=1+$\frac{1}{3}$ + $\frac{1}{3^2}$ +…+$\frac{1}{3^{2021}}$ – $\frac{2022}{3^{2022}}$

   6C=3+1+$\frac{1}{3}$ +…+$\frac{1}{3^{2020}}$ – $\frac{2022}{3^{2021}}$     

   6C-2C=(3+1+$\frac{1}{3}$ +…+$\frac{1}{3^{2020}}$ – $\frac{2022}{3^{2021}}$)-(1+$\frac{1}{3}$ +$\frac{1}{3^2}$ +…+$\frac{1}{3^{2021}}$-$\frac{2022}{3^{2022}}$  

   4C=3-$\frac{2022}{3^{2021}}$+$\frac{2022}{3^{2022}}$

   4C=3-( $\frac{2022}{3^{2021}}$-$\frac{2022}{3^{2022}}$)

   Vì ($\frac{2022}{3^{2021}}$-$\frac{2022}{3^{2022}}$)>0

   =>4C<3

   =>C<$\frac{3}{4}$     

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