Cho dãy số

B

- Hướng dẫn giải

Ta có $x_n=\frac{1}{{\log }_n\left(2010\right)}$ với n = 2;3;4..

Khi đó

$$\begin{gathered} a=x_{11}+x_{12}+x_{13}+x_{14}+x_{24} \\ ={\log }_{2010}\left(11\right)+{\log }_{2010}\left(12\right) \\ +{\log }_{2010}\left(13\right)+{\log }_{2010}\left(14\right) \\ +{\log }_{2010}\left(24\right) \\ ={\log }_{2010}\left(11.12.13.14.24\right) \\ b=x_{63}+x_{64}+x_{65}+x_{66}+x_{67} \\ ={\log }_{2010}\left(63.64.65.66.67\right) \end{gathered}$$

Suy ra

$$\begin{gathered} b-a={\log }_{2010}\left(2.3.5.6.7\right) \\ ={\log }_{2010}\left(2010\right)=1 \end{gathered}$$

Đáp án B