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

Each of Exercises $7-12$ gives the first term or two of a sequence along
with a recursion formula for the remaining terms. Write out the first
ten terms of the sequence.
$$
a_{1}=1, \quad a_{n+1}=a_{n}+\left(1 / 2^{n}\right)
$$

Key Concepts

Recurrence Relation

A recurrence relation is an equation that defines each term of a sequence in relation to preceding term(s). It specifies how the sequence evolves and often depends on one or more initial values. This method of definition is widely used in mathematics, computer science, and related fields to model processes and solve problems where a term’s value is naturally expressed as a function of previous terms.

Iterative Computation of Sequence Terms

This concept involves sequentially calculating the terms of a sequence by repeatedly applying a rule or formula, usually provided by a recurrence relation. Starting from the given initial term(s), each subsequent term is built upon the last, allowing for the step-by-step construction of the sequence. This approach is especially useful when an explicit formula for the nth term is not readily available or is difficult to derive.

Transcript

00:01 We want to find the first 10 terms of the sequence given that a1 is equal to 1 and a n plus 1 is equal to a n plus 1 over 2 to the n so first let's go ahead and let just n equal to 1 so when n is equal to 1 that gives us we'll have a 2 on the left and then we actually better color, so a2, and over here we'll have a 1, and then we'll be plus 1 over 2 to the 1.

00:54 So, a1 was 1 plus 1 half, so that there would be 3 halves, so that will be a 2.

01:09 Now when n is equal to 2, on the left we'll have a is equal to 3, which is equal to a, let me just go ahead and write all the parts that just have n, and then i can go ahead and plug it in, a2 to the second power.

01:34 So using what we found in the last part, plugging that in to here, we'll have three half plus one, fourth, and then adding those together we'll give seven force so we get that a3 is equal to seven force now when n is equal to three the left -hand side is going to be a fourth we'll have a plus one over two and then this should be a third and two to the third now a to the third plus seven course and one over three is one eighth and then adding those two together we get 15 over eight for our fourth term now when n is equal to four we will have a five and this here will be a plus one over two so this would be to the fourth power for a fourth and then to the fourth power from there and a to the fourth was 15 over 8 and then 2 to the fourth is 16 so we'll get 1 plus 16 there and then adding those together we get 31 over 16 for our a to 5 now we can go ahead and do when end is equal to 5.

03:46 So that means our left side will be a6, and then we'll have a plus 1 over 2, and over here on each we should have 5.

04:00 And we found in the last one that a5 is 31 over 16, plus and 2 to the 5th power is 32.

04:11 Now adding those two together, we get 63 over 32, which will be our sixth term in the sequence...

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