The pile of cans described forms an arithmetic sequence where:

• The first layer has $a_1 = 3$ cans,
• The second layer has $a_2 = 6$ cans,
• The third layer has $a_3 = 9$ cans, and so on.

The common difference $d$ between each layer is $6 - 3 = 3$ cans.

To find the total number of cans in the pile when there are 20 layers (terms), we need to:

1. Identify the general formula for the $n$-th term of the sequence.
2. Use the sum formula for an arithmetic sequence.

1. General Formula for the $n$-th Term

The general formula for the $n$-th term of an arithmetic sequence is:

$a_n = a_1 + (n - 1)d$

For this sequence:

$a_n = 3 + (n - 1) \times 3 = 3n$

2. Sum of the First 20 Terms

The sum $S_n$ of the first $n$ terms of an arithmetic sequence is given by:

$S_n = \frac{n}{2} \times (a_1 + a_n)$

Here, $n = 20$, $a_1 = 3$, and $a_{20} = 3 \times 20 = 60$.

So, the total number of cans $S_{20}$ is:

$S_{20} = \frac{20}{2} \times (3 + 60) = 10 \times 63 = 630$

Thus, there are 630 cans in the pile.

Would you like further details or have any other questions?

Related Questions:

1. How would the total number of cans change if the pile had 30 layers instead of 20?
2. What is the number of cans in the 15th layer?
3. If each can weighs 150 grams, what is the total weight of all the cans in the pile?
4. What if the sequence started with 5 cans in the first layer and increased by 5 cans each time—how many cans would be in a 20-layer pile?
5. How would the total number of cans be affected if every second layer had one less can than the sequence predicts?

Tip:

When working with arithmetic sequences, understanding the general term formula is key to solving various related problems efficiently.