6. We set $$\begin{aligned} & x=\sum_{k=0}^{n}\binom{2 n+1}{2 k+1} 2^{3 k}=\frac{1}{\sqrt{8}} \sum_{k=0}^{n}\binom{2 n+1}{2 k+1} \sqrt{8}^{2 k+1}, \\ & y=\sum_{k=0}^{n}\binom{2 n+1}{2 k} 2^{3 k}=\sum_{k=0}^{n}\binom{2 n+1}{2 k} \sqrt{8}^{2 k} \end{aligned}$$ Both $x$ and $y$ are positive integers. Also, from the binomial formula we obtain $$y+x \sqrt{8}=\sum_{i=0}^{2 n+1}\binom{2 n+1}{i} \sqrt{8}^{i}=(1+\sqrt{8})^{2 n+1}$$ and similarly $$y-x \sqrt{8}=(1-\sqrt{8})^{2 n+1} .$$ Multiplying these equalities, we get $y^{2}-8 x^{2}=(1+\sqrt{8})^{2 n+1}(1-\sqrt{8})^{2 n+1}=$ $-7^{2 n+1}$. Reducing modulo 5 gives us $$3 x^{2}-y^{2} \equiv 2^{2 n+1} \equiv 2 \cdot(-1)^{n}$$ Now we see that if $x$ is divisible by 5, then $y^{2} \equiv \pm 2(\bmod 5)$, which is impossible. Therefore $x$ is never divisible by 5. Second solution. Another standard way is considering recurrent formulas. If we set $$x_{m}=\sum_{k}\binom{m}{2 k+1} 8^{k}, \quad y_{m}=\sum_{k}\binom{m}{2 k} 8^{k}$$ then since $\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$, it follows that $x_{m+1}=x_{m}+y_{m}$ and $y_{m+1}=8 x_{m}+y_{m}$; therefore $x_{m+1}=2 x_{m}+7 x_{m-1}$. We need to show that none of $x_{2 n+1}$ are divisible by 5. Considering the sequence $\left\{x_{m}\right\}$ modulo 5, we get that $x_{m}=0,1,2,1,1,4,0,3,1,3,3,2,0,4,3,4,4,1, \ldots$ Zeros occur in the initial position of blocks of length 6, where each subsequent block is obtained by multiplying the previous one by 3 (modulo 5). Consequently, $x_{m}$ is divisible by 5 if and only if $m$ is a multiple of 6, which cannot happen if $m=2 n+1$.