Question

Cho C = \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}.\)Chứng minh rằng C <\(\frac{5}{3}\)

Answer 1

C = \(\frac{5}{4}+\frac{5}{4^2}+\frac{5}{4^3}+...+\frac{5}{4^{99}}\)

= \(5\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)

Đặt A = \(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\)

4A = \(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\)

4A - A = \(\left(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{99}}\right)-\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^{99}}\right)\)

3A = \(1-\frac{1}{4^{99}}< 1\)

=> A < \(\frac{1}{3}\)   (1)

Thay (1) vào C ta được:

\(C< 5\cdot\frac{1}{3}=\frac{5}{3}\)(đpcm)

Answer 2

Ta có:\(\frac{5}{4}\)< \(\frac{5}{3}\)Mà C = \(\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{99}}\)<\(\frac{5}{4}\)

\(\Rightarrow\)C < \(\frac{5}{3}\)

Answer 3

suy luận cùi

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Answer 4

Oxford Dinh lam sai

Answer 5

Cảm ơn nha ST

Answer 6

ST ơi 4A phải=\(1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{98}}\)