# converting a regular summation to a riemann summation

I'm trying to convert the regular summation $$\sum_{i=1}^{30}3i+1$$into a Reimann sum. (and eventually into an integral) The issue, however, is that I can't find a way to format the summation in a way to get $i/n$ into it, such that $\Delta x$ can be extracted and the integral generated. I tried using the formula $$\Delta x = \frac{b-a}{n}\to\frac{30}{n}\to\frac{30}{i}$$ but trying to fit that into the equation only yields something like $$\sum_{i=1}^{30}3\cdot\frac{30}{\Delta x}+1$$ which is nonsensical. I feel like there's a simple solution to this problem but I'm just at the tip. Can anyone offer some hints as to where I should try next?

note: my problem isn't converting a riemann sum into an integral; this is about converting a regular sum into a reimann sum, something I haven't found much help for on StackExchange.

• Just for the typos: In the title "Riemann", not riemann; in the text "Riemann", not Reimann. Lat two lines: replace "riemann" and "reimann" as well... – Dietrich Burde Sep 16 '18 at 16:07