6. (8 points) By expanding the expression $(1+\sqrt{13})^{210}$ using the binomial theorem, we obtain terms of the form $C_{210}^{k}(\sqrt{13})^{k}$. Find the value of $k$ for which such a term attains its maximum value.
Answer: 165

Solution: The ratio of two consecutive terms $\frac{C_{210}^{k+1}(\sqrt{13})^{k+1}}{C_{210}^{k}(\sqrt{13})^{k}}$ is greater than 1 when $k<$ $\frac{210 \sqrt{13}-1}{\sqrt{13}+1}$. Then the terms increase up to $\left[\frac{210 \sqrt{13}-1}{\sqrt{13}+1}\right]+1$, and then decrease.