Cho A= 7 + 7^2 + 7^3 + 7^4+ ... + 7^99+ 7^100. Tìm x để biết 6A + 7 = 7^(2x + 1)

Cho  \(A{\rm{ }} = {\rm{  }}7 + {7^2} + {7^3} + {7^4} + ... + {7^{99}} + {7^{100}}{\rm{ }}\).

 Tìm \(x\) để biết \(6A + 7 = {7^{2x + 1}}\)

Ta có:

\(A{\rm{ }} = {\rm{  }}7 + {7^2} + {7^3} + {7^4} + ... + {7^{99}} + {7^{100}}{\rm{ }}\)

\(7A{\rm{ }} = {\rm{ }}{7^2} + {7^3} + {7^4} + ... + {7^{99}} + {7^{100}}{\rm{  + }}{{\rm{7}}^{101}}\)

\(\begin{array}{l}7A{\rm{  -  A}} = \left( {{\rm{ }}{7^2} + {7^3} + {7^4} + ... + {7^{99}} + {7^{100}}{\rm{  + }}{{\rm{7}}^{101}}} \right) - {\rm{ }}\left( {7 + {7^2} + {7^3} + {7^4} + ... + {7^{99}} + {7^{100}}{\rm{ }}} \right)\\6A = {7^{101}} - 7\end{array}\)

\(\begin{array}{l}{7^{101}} - 7 + 7 = {7^{2x + 1}}\\{7^{2x + 1}} = {7^{101}}\\2x + 1 = 101\\x = 50\end{array}\)