# Can somone help me do this double sum problem. I know how to do it manually, but I would like to know how to do it using summation formulas.

Calculating the double sum: $$\sum\limits_{i=1}^{10}\sum\limits_{j=0}^{15}(3i+2j)$$

I know how to do this manually, but I would like to know how to do it using a summation formula. Could somone please show me the steps as to how I would be able to solve this using summation formulas?

\begin{align*} \sum_{i=1}^{10} \sum_{j=0}^{15} (3i + 2j) &= \sum_{i=1}^{10} \left( (15-0+1)(3i) + \sum_{j=0}^{15} 2j \right) \\ &= \sum_{i=1}^{10} \left( 48i + 2 \sum_{j=0}^{15} j \right) \\ &= 48 \sum_{i=1}^{10} i + 2(10 - 1 + 1) \sum_{j=0}^{15} j \\ &= 48 \frac{10(10+1)}{2} + 20 \frac{15(15+1)}{2}. \end{align*}
Let's look at $\sum_{j=0}^{15}(3i+2j)$ where $i$ is some number unknown to us.
This is an arithmetic series where $a_0=3i$, $a_{15}=3i+30$, and we have $16$ elements overall.
So that summation according to gauss is $\frac{16(3i+3i+30)}{2} = 48i+240$.
You now need to compute $\sum_{i=1}^{10}(48i+240)$. Can you take it from here?
Reminder: The sum of an arithmetic series of $n$ numbers, where the first is $a_0$ and the last is $a_{n-1}$ is $S=\frac{n(a_0+a_{n-1})}{2}$