Question

Chứng minh rằng : \(\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}< \frac{1}{50}\)

Answer 1

M = 512 - 512/2 - .... - 512/2^10
   = 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
   = 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1 
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9  + 2^8 + 2^7 +... +    1 + 1/2
          M   = 2^10 - 2.2^9 + 1/2
          M  = 2^10 - 2^10 + 1/2
          M  = 1/2

Answer 2

Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{4n-2}}-\frac{1}{7^{4n}}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)

\(\Rightarrow49A=1-\frac{1}{7^2}+...+\frac{1}{7^{4n-4}}-\frac{1}{7^{4n}}+..+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)

\(\Rightarrow49A+A=50A=1-\frac{1}{7^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{50}=\frac{1}{50}-\frac{1}{7^{100}.50}< \frac{1}{50}\left(ĐPCM\right)\)

Answer 3

M = 1/2 nha

Answer 4

Gọi A=172−174+176−...+1798−17100A=721​−741​+761​−...+7981​−71001​

49A=1−172+174−...+1796−179849A=1−721​+741​−...+7961​−7981​

49A+A=(1−172+174−...+1796−1798)+(172−174+176−...+1798−17100)49A+A=(1−721​+741​−...+7961​−7981​)+(721​−741​+761​−...+7981​−71001​)

50A=1−1710050A=1−71001​

A=1−1710050<150A=501−71001​​<501​

Vậy A<150A<501​