### Assignments:

Unfinished Assignment Study Questions for Lesson 33

### Lesson Objectives:

- Find the terms of a sequence given its nth term.
- Determine a general term for a sequence.
- Convert between sigma notation and other notation for a series.
- Construct the terms of a recursively defined sequence.

An infinite sequence is a function whose domain is made up of an infinite set of positive integers.

Whereas a finite sequence is a function whose domain is made up of a finite set of positive integers.

Given the nth term of a sequence, find the first four terms, the tenth term, and the 15th term.

The nth term, denoted a_n, is 5n-1.

The first four terms of our sequence, are a_1, a_2, a_3, and a_4.

a_1 = 5(1)-1. This is 4.
a_2 = 5(2)-1. This is 9.
a_3 = 5(3)-1. This is 14.
a_4 = 5(4)-1. This is 19.

So the first four terms of our sequence are 4, 9, 14, and 19.

Now our 10th term, a_10, is equal to 5(10) - 1, which is 49. And our 15th term, a_15, is equal to 5(15) - 1. This is 74.

Predict the general term, or the nth term, a_n, of the sequence.

5, 10, 15, 20, 25

We need to look for a pattern in the sequence, and determine a general term.

For our first term, n = 1, 5*1 = 5. For our second term, n = 2, 5*2 = 10. For our third term, n = 3, 5*3 = 15. For our fourth term, n = 4, and 5*4 = 20. And then for our fifth term, n = 5, 5*5 is equal to 25.

So our general term is a_n = 5*n.

Sums and Series.

The sum of the terms in an infinite sequence is called an infinite series.

The sum of the first n terms is called the nth partial sum, or a finite series, denoted S_n.

Find the indicated partial sums for the sequence.

S_3 and S_7.

We want to find the 3rd partial sum, and the 7th partial sum for the sequence: 2, 4, 6, 8, 10, 12, 14, etc..

So the 3rd partial sum is equal to the first 3 terms added together. 2+4+6. This is equal to 12.

The 7th partial sum is equal to the first 7 terms added together. 2+4+6+8+10+12+14. This is equal to 56.

Find and evaluate the sum.

This is read, "the sum as k goes from 1 to 5 of 1 over 3k." And this is the greek letter sigma, which denotes a sum.

We replace k with 1, 2, 3, 4, and 5, because k goes from 1 to 5. And then add the results.

So we have 1/(3*1)+1/(3*2)+1/(3*3)+1/(3*4)+1/(3*5). So we have 1/3+1/6+1/9+1/12+1/15. This equals 137/180.

Write the sum in sigma notation. Answers may vary.

7+14+21+28+35+42+49, etc.

So here we have an infinite series, and we have multiples of 7. That means when n = 1, our first term, 7(n), or 7(1) = 7. Then for our second term, when n = 2, 7(n), or 7(2) = 14. Then for our third term, when n = 3, 7(n), or 7(3) = 21. And this applies to every term in our series. 7(4) = 28, 7(5) = 35, 7(6) = 42, and 7(7) = 49.

So our general term is 7n. So in sigma notation, we have the sum as n goes from 1 to infinity of 7n.

A recursively defined sequence lists the first term or terms and a recursive formula, which uses the preceding term (or terms) to define the remaining terms.

Find the first 4 terms of the recursively defined sequence.

We're given our first term, a_1 = 2, and we're given the recursive formula, a_(n+1) = 2+1/(a_n).

Since we're already given our first term, we need to find our second, our third, and our fourth terms.

From our recursive formula, we have a_2, when n = 1, since 1+1 gives us 2. For our second term, we have 2+1/("our first term").

For our third term, we have n = 2. So, 2+1/("second term").

For our fourth term, we have n = 3, so 2+1/("third term").

Now let's evaluate our second term. This is 2+1/2. This equals 5/2.

Then let's evaluate our third term. This is 2+1/(5/2), or 2+2/5, which is 12/5.

Then for our fourth term, we have 2+1/(12/5), or 2+5/12, which is 29/12.

So the first four terms of our sequence are 2, 5/2, 12/5, and 29/12.