### 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`.