Cho hàm số f(x) có f(1) = 1, f(m+n) = f(m) + f(n) + m

* Hướng dẫn giải

Đáp án B.

Cho m=1 ta có

$\begin{array}{l}f(n+1)=f\left(n\right)+f\left(1\right)+n\\ \iff f(n+1)=f\left(n\right)+n+1.\end{array}$ 

Khi đó 

$\begin{array}{l}f\left(2\right)+f\left(3\right)+\mathrm{...}+f\left(k\right)\\ =\left(f\left(1\right)+2\right)+\left(f\left(2\right)+3\right)\\ +\mathrm{...}+\left(f(k-1)+k+1\right)\end{array}$

 $\begin{array}{l}\iff \left[f\left(2\right)+f\left(3\right)+\mathrm{...}+f(k-1)\right]+f\left(k\right)\\ =f\left(1\right)+\left[f\left(2\right)+\mathrm{...}+f(k-1)\right]+(1+2+\mathrm{...}+k)\end{array}$ 

$\begin{array}{l}\iff f\left(k\right)=f\left(1\right)+(1+2+\mathrm{...}+k)\\ =1+\frac{k(k+1)}{2}.\end{array}$

Vậy hàm cần tìm là 

$\begin{array}{l}f\left(x\right)=1+\frac{x(x+1)}{2}\\ \Rightarrow \left\{\begin{array}{l}f\left(96\right)=1+\frac{96.97}{2}=4657\\ f\left(69\right)=1+\frac{69.70}{2}=2416\end{array}\right.\end{array}$

Vậy

 $\begin{array}{l}T=\log \frac{4657-2416-241}{2}\\ =\log 1000=3.\end{array}$