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D

- Hướng dẫn giải

Ta có:

$$\begin{gathered} \frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} \\ =6\left(\frac{3^k-2^k}{3^{k+1}-2^{k+1}}-\frac{3^{k+1}-2^{k+1}}{3^k-2^k}\right) \\ \sum _{k=1}^n\frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} \\ =6\frac{3^n-2^n}{3^{n+1}-2^{n+1}} \end{gathered}$$

Do đó:

$$\begin{gathered} \underset{x\to \infty }{\lim }\sum _{k=1}^n\frac{6^k}{\left(3^{k+1}-2^{k+1}\right)\left(3^k-2^k\right)} \\ =6\underset{n\to \infty }{\lim }\frac{\left(3^n-2^n\right)}{3^{n+1}-2^{n+1}} \\ =6\underset{n\to \infty }{\lim }\frac{1-{\left({\displaystyle \frac{2}{3}}\right)}^n}{1-2.{\left(\frac{2}{3}\right)}^2}=2 \end{gathered}$$

Đáp án D