Cho \(f\left( 1 \right) = 1,f\left( {m + n} \right) = f\left( m \right) + f\left( n \right) + mn\) với mọi \(mn \in {N^*}\).

* Hướng dẫn giải

\(\begin{array}{l}
n = 1 \Rightarrow f\left( {m + 1} \right) = f\left( m \right) + f\left( 1 \right) + m.1 \Leftrightarrow f\left( {m + 1} \right) = f\left( m \right) + m + 1\\
 \Leftrightarrow f\left( {m + 1} \right) - f\left( m \right) = m + 1\\
 \Rightarrow f\left( {96} \right) - f\left( {69} \right) = \left[ {f\left( {96} \right) - f\left( {95} \right)} \right] + \left[ {f\left( {95} \right) - f\left( {94} \right)} \right] + ... + \left[ {f\left( {70} \right) - f\left( {69} \right)} \right]\\
 = 96 + 95 + ... + 70 = \frac{{27.\left( {96 + 70} \right)}}{2} = 2241\\
 \Rightarrow \frac{{f\left( {96} \right) - f\left( {69} \right) - 241}}{2} = \frac{{2241 - 241}}{2} = 1000\\
 \Rightarrow T = \log \left[ {\frac{{f\left( {96} \right) - f\left( {69} \right) - 241}}{2}} \right] = \log 1000 = 2.
\end{array}\)