Question

[Nguyễn Hữu Thế]

So sánh: A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+..........+\frac{1}{7^{100}}\)

               \(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+.........+\frac{1}{7^n}+\frac{1}{7^{n+1}}\)

               \(C=\frac{7}{41}\)

Hạo ơi ông giúp t vs! Trên lp thầy có giảng rồi nhưng quên oày! 

Answer 1

[soyeon_Tiểubàng giải]

 \(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)

\(7A=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)

\(7A-A=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)

\(6A=1-\frac{1}{7^{100}}< 1\)

\(A< \frac{1}{6}=\frac{7}{42}< \frac{7}{41}=C\)

=> \(A< C\)

\(B=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^n}+\frac{1}{7^{n+1}}\)

\(7B=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{n-1}}+\frac{1}{7^n}\)

\(7B-B=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{n-1}}+\frac{1}{7^n}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^n}+\frac{1}{7^{n+1}}\right)\)

\(6B=1-\frac{1}{7^{n+1}}< 1\)

\(B< \frac{1}{6}=\frac{7}{42}< \frac{7}{41}=C\)

Answer 2

[Kẹo dẻo]

Nguyễn Hữu Thế fai gọi bằng cách này này:

Hạo ơi giúp vs.

Vậy Lê Nguyên Hạo ms nhận đc thông báo.