$\sum_{i=1}^n \left(\frac{14i}{n}-5\right)\frac{4}{n}$ and where $\sum_{i=1}^n i=1+2+3+\cdots+n=\frac{n(n+1)}{2}$
Not entirely sure if I did this correctly, but I basically plugged in $\frac{n(n+1)}{2}$ for $i$ and after some algebra and simplification I got $\sum_{i=1}^{n} \frac{28n+8}{n}$
The reason your answer was incorrect is because you replaced $i$ with $\frac{n(n+1)}2$, but $i$ is not equal to that. The sum of all integers up to a number $n$ is equal to that. So the goal is to get $i$ in a position where it represents the sum of all integers up to $n$ $$\sum_{i=1}^n\left(\frac{14i}n-5\right)\frac4n=\frac4n\left(\sum_{i=1}^n\frac{14}n \cdot i-\sum_{i=1}^n5\right)$$ $$=\frac4n\left(\frac{14}n\sum_{i=1}^ni-5n\right)=\frac4n\left(\frac{14}n \cdot \frac{n(n+1)}2-5n\right)$$ $$=\frac4n(2n+7)=\frac{8n+28}n$$ I probably went through more detail than needed, but hopefully it was effective. Notice how I didn't substitute anything for $\frac{n(n+1)}2$ until there was nothing else in the sum but $i$.