How is the following arithmetic sequence solved?

Question

Apologies to bother you with this, but how is the following arithmetic sequence solved?

$$\dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)$$

Answer 1

Write $$ \dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)=\dfrac{n}{2n}\sum_{k=1}^{n-1}1-\dfrac1{2n}\sum_{k=1}^{n-1}k+\dfrac1{2n}\sum_{k=1}^{n-1}1 $$ and see this.

Answer 2

First, you can pull out everything from the sum that does not depend on $k$. So $$\dfrac1n \left(\sum_{k=1}^{n-1}\dfrac{n-k+1}2\right)=\frac 1n \left(\frac {(n-1)(n+1)}2-\frac 12\sum_{k=1}^{n-1}k\right)$$ where I pulled out $\frac {n+1}2$ and multiplied by $n-1$ as the number of terms. Can you do the last?