College Algebra Tutorial 54

College Algebra
Answer/Discussion to Practice Problems
Tutorial 54A: Sequences

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Answer/Discussion to 1a

problem 1a ; 1 < n < 5

Basically, to find the nth term of a sequence works in the same fashion as function notation. If you want to find the 3rd term, you are looking for example 1b

, which means you plug in 3 for n into the given function.

So, what are we going to plug in for n to find the 1st term? If you said 1, give yourself a pat on the back. What about the 2nd term? I hope you said you would plug in 2 for n.

Since we have to go from 1 < n < 5, this means we need to find 5 terms and we will be plugging in 1, 2, 3, 4, and 5 for n.

Let's see what we get:

*1st term, n = 1

*2nd term, n = 2

*3rd term, n = 3

*4th term, n = 4

*5th term, n = 5

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Note how we had -1 raised to n, which changes value, and the signs of the terms alternated.

The five terms of this sequence are 1/4, -1/8, 1/16, -1/32, and 1/64.

(return to problem 1a)

Answer/Discussion to 2a

problem 2a

This function contains a factorial.

Let's see what we get for our first five terms:

problem 2a

*1st term, n = 1

*2nd term, n = 2

*3rd term, n = 3

*4th term, n = 4

*5th term, n = 5

Now let's check out the twelfth term:

*12th term, n = 12

(return to problem 2a)

Answer/Discussion to 3a

problem 3a

Let's take a look at what is happening here:

Something that is always constant is that each term contains e.

There are also two things that change.

First let's look at the alternating signs:

For it to have alternating signs, we need to have (-1) raised to a power that changes. This means n, the term number is involved.

The first term is positive, the second term is negative, the third positive, the fourth negative and so forth.

When n is odd (1, 3, 5, ...), then the term is positive.

When n is even (2, 4, 6, ...), then the term is negative.

So do you think we are going to have example 5a

.

If you said example 5b

you are correct!!! If n is odd, then this term will be positive. If n is even, then this term will be negative.

We also have the exponent on each term.

Again we need to figure out the relationship between n and the exponent:

When n is 1, the exponent is 1.
When n is 2, the exponent is 4.
When n is 3, the exponent is 9.
When n is 4, the exponent is 16.

What do you think the relationship is?
It looks like the exponent is always n squared.

Putting it together, the formula for the nth term is ad3a3

.

Sometimes you have to play around with it before you get it just right. You can always check it by putting in the n values and seeing if you get the given sequence.

This one does check.

(return to problem 3a)

Answer/Discussion to 4a

problem 4a ; problem 4a2

We are giving the first term, problem 4a2

. Using that we can find the second term and so forth.

Let's see what we get for our first three terms:

*2nd term, n = 2

*3rd term, n = 3

Since this is a recursive formula, in order to the fifth term, we need to find the fourth term:

*4th term, n = 4

*5th term, n = 5

(return to problem 4a)

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