Each of Exercises 7-12 gives the first term or two of a sequence along with a recursion formula

Each of Exercises 7-12 gives the first term or two of a...

Key Concepts

Alternating Sign Factor

When a recurrence relation includes an alternating sign factor, such as one raised to a function of the term index, it causes the terms of the sequence to switch signs in a regular pattern. Understanding how this factor affects the sequence is important for predicting its behavior.

Iteration

Iteration in the context of recursive sequences refers to the process of repeatedly applying the recurrence relation to compute the next term from the current or previous ones. This iterative process is fundamental to constructing the sequence step by step.

Recursive Sequences

A recursive sequence is defined by specifying one or more initial terms and a recurrence relation that defines each subsequent term based on preceding ones. This approach emphasizes the step-by-step construction of the sequence rather than providing a direct formula for the nth term.

Recurrence Relation

A recurrence relation is a mathematical expression that relates a term in a sequence to one or more of its earlier terms. This relation serves as a rule for iteratively generating the sequence once the initial conditions are known.

Initial Conditions

Initial conditions establish the starting points of a sequence from which all other terms are generated using the recurrence relation. They are crucial because they uniquely determine the sequence.

Transcript

00:01 We want to find the first 10 terms of the sequence given a 1 is equal to 2 and a n plus 1 is equal to negative 1 to the n plus 1 times a n all over 2.

00:16 So when n is equal to 1, well, we plug in 1 on the left.

00:26 So we'd get a 2 is equal to 2, well, it's going to be negative 1 to the first power.

00:37 Plus 1 and then a1 and then all this is divided by 2.

00:46 So we're just told that a1 is 2 so we can plug that in.

00:50 So we get a 2 there and then negative 1 to the second power is just 1.

00:56 And then we would divide that to just get 1 or the second term in our sequence.

01:07 Then when n is equal to 2, the left side will become a 3 and then we'll have negative 1 to the second plus 1 a 2 all over 2 and then we take what we found for a 2 plug that in and then negative 1 to the third power will be negative 1 all over 2 so we end up with negative 1 half for the third term in our sequence next we'll have n is equal to 3.

01:52 So the left -time side will have a4 is equal to so negative 1 now to the 3rd plus 1 and then we'll have a3 and then still all divided by 2.

02:09 So take a3, plug that in, negative 1 1⁄2.

02:14 Negative 1 to the 4th would just be 1 divide that by 2.

02:20 So we end up with negative 1 4th.

02:25 For our fourth term then when n is equal to not four i mean not five but four we will have a5 is equal to negative one to the four plus one and then a four and we still divide all this by two so negative one to the fifth will be negative one a four was negative 1 fourth, divide that by 2.

03:11 So the negatives cancel out, and we'll be left with 1 .4th, divide that by 2, which gives us 1 .8th.

03:17 So this is our fifth term in the sequence.

03:21 Now, when n is equal to 5, it's going to be the same thing.

03:25 So a6 on the left is equal to, so negative 1, now to the 5th, plus 1, a5, and then we divide all that by 2.

03:41 Take the 1 eighth, plug it in, and so negative 1 to the 6 is just 1.

03:48 A5 is 1 8th, divide that by 2, so we end up with 1 over 16 for our 6 term in the sequence.

03:57 Now when n is equal to 6, so a7 is equal to, so negative 1 to the 6 plus 1, and then a7 is equal 2, so negative 1 to the 6 plus 1, and then a7, 6 all over 2...

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