Can a summation be expressed in terms of two variables? How?

Question

Given the following question. What is the sum of all factors of 72?

72 can be expressed as 2^3 x 3^2. The following are the factors of 72. 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72 You simply add all.

However I have noticed that you can conceptualize the sum of the factors as simply a summation based on two variables. It is the summation of 2^x times 3^y where x is from 0 to 3 and y is form 0 to 2.

Because x is from 0 to 3, there are 3 + 1 or 4 options. Because y is from 0 to 2, there are 2 + 1 or 3 options. Multiplying the combination of these two options through fundamental principle of counting, you get 12 factors.

My question is, is there a formal summation representation in terms of two variables to account for the concept I have stated? I have trouble searching in the internet since I don't know the correct terms to put.

Also, I apologize for not being able to use LateX since I am currently on mobile.

Answer 1

"What is the sum of all factors of 72"

$\sum_{d|72} d$.

"My question is, is there a formal summation representation in terms of two variables to account for the concept I have stated?"

$\sum_{x=0...3;y=0...2}2^x3^y$ or

$\sum_{x\in \mathbb Z, 0\le x \le 3, y\in \mathbb Z, 0\le x\le 2} 2^x3^y$ of

Or $\sum_{(x,y)\in \mathbb Z_4\times \mathbb Z_3}2^x3^y$

You can also do a double sum $\sum_{x=0}^3\sum_{y=0}^2 2^x3^y$ which is intrepeted as $\sum_{x=0}^3 (\sum_{y=0}^2 2^x3^y)$.

You can do $\sum_{\text{any describable property for the index value or values}} whatever$

$\sum_{k=a}^b s_k$ is just shorthand, but very common, shorthand for $\sum_{k\in \{a,a+1, ......, b-1, b\}} s_n$.

Note: $\sum_{w\in S} w$ where $S$ is a set of elements is also very common.

.......

BTW is calculating $\sum_{d|72}d$ it might be useful to note that the sum equal $\sum_{x=0}^3\sum_{y=0}^2 2^x3^y=\sum_{x=0}^3 (2^x\cdot \sum_{y=0}^23^x)$ so that the answer is

$1+3+9+2+6+18+4+12 +36 + 8+ 24 + 72 =$

$(1+3+9) + 2(1+3+9) + 4(1+3+9) + 8(1+3+9)=$

$(1+2+4+8)(1+3+9) = $

$(2^4 -1)(\frac {2^3 -1}{3-1}) = 17\times \frac {26}2=17\times 13$.

But you weren't asking about that, were you?

Answer 2

You can chain summations:

$$\sigma = \sum_{k_2=0}^3\sum_{k_3=0}^2 2^{k_2}3^{k_3}$$

You can also use set notation to describe your pile and then sum that

$$S=\left\{2^{k_2}3^{k_3}\middle|0\le k_2\le 3\in\mathbb N,0\le k_3\le 2\in\mathbb N\right\}$$ $$\sigma = \sum_{n\in S}n$$