![]() ![]() Since d is common for all the equations, it is easier to use and remove d. Step 3: The reason we use simultaneous equations is so we can remove one of the terms. We can use simultaneous equations for this as well, but it is a bit different from what we did before. Step 2: Once again, we can substitute the n values for 1, 2, and 3 this time because we need to have 3 variables and 3 unknowns to find out three particular equations. ![]() Since the 3rd difference = 6 a, the value of a = 1. We can see here that the 3rd difference is 6. We write down the n value, then the term value, and then we calculate not only the first difference, the second difference but the 3rd difference as well. Step 1: We repeat what we did in the previous sequences. Step 6: Finally, substituting all the values we have found for a, b, and c, we can give the n th term in the equation form:Ĭubic sequences have a constant third difference. In this case since both of these expressions have + c, we can just subtract both equations. Step 5: Use simultaneous equations to solve these two equations. We substitute the value of a as 1, and we get two equationsĮquation 1 u n = n 2 + b n + c where n = 1,Įquation 2 u n = n 2 + b n + c where n = 2, Therefore, we try to form two equations with 2 unknowns using the value of a. Step 4: Using the value of 1, we can substitute it into the common expression as u n = n 2 + b n + c, but we still need to work out the value of b and c. Therefore 2 a = +2 (we found this out in the earlier step), and from here, we can work out the value of a = 1. Step 3: Remember we said that 2 a = the second difference. This means that the difference between two consecutive numbers here is +2. In this example we can see that the 1st difference between the numbers is +5, +7, +9 and +11. The second difference is the difference between consecutive numbers in the first difference (read that carefully, it might be a bit confusing!) ![]() Step 2: Once the first difference is calculated, the second difference also needs to be calculated. In this case, we can see that the difference between the numbers is +5, +7, +9 and +11. ![]() Step 1: Just like last time, we can put these values in a table and calculate the first difference between the numbers. We can consider an example sequence: 2, 7,14, 23, 34. In quadratic sequences, what must be known is that the second difference is going to be equal to 2 a. Whereas in linear sequences, the first difference is always the same, in quadratic sequences, it is the second difference which is always the same. So hence the final equation is u n = 4 n – 2. In this example, it can be seen that we need to add 2 every time. Step 4: Then we need to compare both the columns 4 n and the value of a to see how much more we need to add to the 4 n value for a to be the answer. Therefore, the 4 n values for n = 1, 2, 3, 4, will be 4, 8, 12, and 16 respectively. This means we are multiplying the n value here by 4. In order to do that, we can introduce another column to your table which is 4 n. Step 3: Now that we have worked out the value of a, we have to work out the value of b. In this example, we can see that you add 4 to the previous number to get the new number (6 + 4 = 10, 10 + 4 = 14, etc). Step 3: The next thing we have to see is the difference between each of the terms. Step 2: Next to that we have to write each of the terms given as 6, 10, 14 and 18. We know that these numbers given above are the first four terms of the sequence, so we write n = 1, 2, 3, 4. Step 1: We put down the terms in a table. Let’s look at a more detailed example of what this is.Įx: If we have the sequence 6, 10, 14, 18 and we are required to obtain the n th term, what is the process that we must follow? Linear SequencesĪ general linear sequence has the general form of u n = a n + b, where a and b are constants and u n is the n th term. ![]()
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