Ta có:

Đặt A=\(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{50}}\)

⇒7A=\(\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{51}}\)

⇒7A-A=\(\frac{1}{7^{51}}-\frac{1}{7}\)

⇒6A=\(\frac{1}{7^{51}}-\frac{1}{7}\)⇒A=\(\frac{1}{6.7^{51}}-\frac{1}{6.7}\)

⇒C=\(\frac{1}{6.7^{51}}-\frac{1}{6.7}\)+\(\frac{1}{6.7^{50}}\)

=\(\frac{4}{3.7^{51}}-\frac{1}{42}\)

\(7^50\) là cái gì????????

\(A=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^5}\)

\(\Rightarrow7A=1+\frac{1}{7}+...+\frac{1}{7^4}\)

\(\Rightarrow7A-A=1-\frac{1}{7^5}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^5}}{6}\)

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

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

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

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

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

khó thể xem trên mạng