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This is a long post, so I've italicized questions and bolded answers. Let me know if there are any mistakes.

The first thing to realize is that we are dealing with finite arithmetic series, that is the sum of the numbers in an arithmetic sequence. We know that each pile will always have 1 can at the top, then 2 cans beneath that, 3 cans beneath that, and so on. It's the bottom layer that varies from question to question. So far, we know that for each pile, the first term in the sum will always be 1 (the top layer). The common difference will be 1, because the number of cans increases by one for each additional layer below. If there's one layer, there's one can total. If there's two layers, there's 1+2 = 3 cans total. If there's three layers, there's 1+2+3 = 6 cans total.

Given this information, how can we find the total number of cans for an arbitrary number of layers, n, which is a positive integer (1, 2, 3, 4...)? First, let the sum be denoted by S. For a sum of n layers, the sum can be represented by:

(1)      S = 1 + 2 + 3 +...+ (n-2) + (n-1) + n

but it can also be rearranged:

(2)      S = n + (n-1) + (n-2) +...+ 3 + 2 + 1

Here's the trick to evaluating the sum. Adding the two equations, notice that when you line each term up in the two equations, their individual sums add up to the same quantity. The first term in equation 1 is 1, and the first term in equation 2 is n. Their sum is n+1. Let's look at the second term in both equations. (n-1) + 2 = n + 1. The third term: (n-2) + 3 = n+1. For the last term, it's the same: 1 + n = n+1. When we add the two equations, we get:

(3) 2S = (n+1) + (n+1) + (n+1) +...+ (n+1) + (n+1) + (n+1).

What does equation 3 evaluate to? Well, there are n terms to add together, and all those terms are equal to each other, so we should multiply (n+1) by an n number of terms: (n+1) * n = n(n+1)

(4) 2S = n(n+1)

Divide both sides by two

(5) S = n(n+1)/2

Remember, S is the total number of cans, and n is the number of layers. This equation applies for a pile of cans where the top layer has one can and each subsequent layer below has one additional can than the layer above. Finally, we have found a formula that generalizes the number of cans in an arbitrary number of layers using Equation (5).

Now, we can answer the questions.

13a: The last term is 20. Notice in Equation (1), the last term is n, so n=20 in this question. Here, we should also note that the number of cans in the bottom layer also equals the number of layers. Using Equation (5), S = 20*21/2 = 10*21 = 210. This matches with the number we were supposed to find

13b. Instead of the number of layers, we are given the total number of cans, which is S. S = 3240. n represents the number of layers as well as the number of cans in the bottom row, so we should solve for n. using Equation (5). Substitute 3240 for S and solve:

3240 = n(n+1)/2

6480 = n^2 + n

0 = n^2 + n - 6480.

The following equation is a quadratic. Using the quadratic formula, we can find n. n = (-1 +/- sqrt(1- 4*1*(-6480)))/2 =

(-1 +/- sqrt(25921))/2. This leads to n = 80 and n = -81. Only the positive answer makes sense, so we choose n = 80 as the number of cans in the bottom row.

13c i: Let's solve n^2 + n - 2S = 0 for S.

n^2 + n = 2S

2S = n^2 +n

S = (n^2 + n)/2. Then factor out an n

S = n(n+1)/2. We've already showed how to derive this equation, so we actually answered this question in the beginning.

13c ii: Let's try to solve it similarly to 13b. Skipping a few steps, we should get n^2 + n - 4200 = 0. The quadratic formula gives us n = (-1 +/- sqrt(1 - 4*1*(-4200)))/2 = (-1 +/- sqrt(16801))/2. However, sqrt(16801) is about 129.62. n will not turn out to be a positive integer for either answer, so there is no pile of cans that has 2100 cans exactly.

14a: If S_n = u_1 + u_2 + u_3 + ... + u_n, and n = 1, then S_1 should only sum up the first term. S_1 = u_1 = 7

14b: S_1 = u_1, and S_2 = u_1 + u_2. Solve for u_2:

S_2 - S_1 = (u_1 + u_2) - (u_1) = u_2 = 18 - 7 = 11. The common difference should be the difference between each consecutive term in the arithmetic series e.g. u_2 - u_1 = 11 - 7 = 4. The common difference is 4.

14c: All we have to do is determine the first four terms in the series. Each subsequent term should be 4 greater than the previous one. The first term is 7 and the second term is 11. Add 4 to get 15 for the third term. Add 4 to get 19 for the fourth term. 19 is the fourth term, or u_4.