6、Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\frac{(n+1) a_{n}}{2 n+a_{n}}\left(n \in N_{+}\right)$. Then $\sum_{k=1}^{2017} \frac{k}{a_{k}}=$
6. $2^{2018}-2019$

Analysis: From $a_{n+1}=\frac{(n+1) a_{n}}{2 n+a_{n}}$, we know that $\frac{n+1}{a_{n+1}}=\frac{2 n}{a_{n}}+1$. Let $b_{n}=\frac{n}{a_{n}}$, then we have $b_{n+1}=2 b_{n}+1$. It is easy to see that $b_{n}=2^{n}-1$, so $\sum_{k=1}^{2017} \frac{k}{a_{k}}=\sum_{k=1}^{2017} b_{k}=2^{2018}-2019$