Biết \({\log _2}\left( {\sum\limits_{k = 1}^{100} {\left( {k \times {2^k}} \right)}  - 2} \right) = a + {\log _c}b\) với \(a,b,c\) 

* Hướng dẫn giải

\(\begin{array}{l}
\sum\limits_{k = 1}^{100} {\left( {k{{.2}^k}} \right) = \left( {2 + {2^2} + ... + {2^{100}}} \right) + \left( {{2^2} + {2^3} + ... + {2^{100}}} \right) + ... + {2^{100}}} \\
 = 2\left( {{2^{100}} - 1} \right) + {2^2}\left( {{2^{99}} - 1} \right) + ... + {2^{99}}({2^2} - 1) + {2^{100}}(2 - 1)\\
 = {100.2^{101}} - \left( {2 + {2^2} + ... + {2^{100}}} \right) = {100.2^{101}} - 2\left( {{2^{100}} - 1} \right)\\
 = {99.2^{101}} + 2\\
 \Rightarrow {\log _2}\left( {\sum\limits_{k = 1}^{100} {\left( {k{{.2}^k}} \right)} } \right) = {\log _2}\left( {{{99.2}^{101}}} \right) = 101 + {\log _2}99\\
 \Rightarrow a = 101,b = 99,c = 2 \Rightarrow a + b + c = 202
\end{array}\)