Kellie writes: “I am told to find the nth number in a pattern. What the heck does nth mean?”

So I’ll first answer the question directly. Then we can talk about what it means to the rest of the problems.

One more piece of the puzzle before we’re on to solve problems. I’ll bet you saw something in there about *n-1* or *n+1*. Here’s a bit about that…

Now let’s get into writing the *n*th term of a sequence using the one before it. Remember the *n*th term is the number in the list if you count up to whatever you’ve labeled *n*.

In the previous video, we would have written it in fancy terms like this:

Once more, just to get the hang of it:

Again, this one would have been written:

Aha! But what if you don’t know the number *right before* the one you want? Watch this:

Still have questions? Post them in the comments. Need something else? Request it and I’ll get to it as soon as I can.

Thanks Kellie, for your question!

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Hi Bon,

Have had lots of questions, just real busy and so I am faking it till I can make it.

How do I know if a set is closed under the rule of addition, same question for multiplication? Thought I understood it, guess I was wrong because I missed all 5 problems on a test.

Thanks for your help.

Thanks, Kellie! I’ve responded with this post.

Hi I got a question. I’m trying to find the recursive function for this sequence 1,7,21,43,73,111,157,211 but can’t seem to get it. How do I go about?

I’m going to type out my thinking as I go, so you can watch it happen…

1->7 (difference of 6)

7->21 (diff of 14)

21->43 (diff of 22)

43->73 (diff of 30)

73->111 (diff of 38)

111->157 (diff of 46)

157->211 (diff of 54)

Hmm…

1->7 (difference of 6=2*3)

7->21 (diff of 14=2*7)

21->43 (diff of 22=2*11)

43->73 (diff of 30=2*3*5)

73->111 (diff of 38=2*19)

111->157 (diff of 46=2*23)

157->211 (diff of 54=2*3*3*3)

Hmm…

1->7 (difference of 6=2*3)

7->21 (diff of 14=2*7)

21->43 (diff of 22=2*11)

43->73 (diff of 30=2*15)

73->111 (diff of 38=2*19)

111->157 (diff of 46=2*23)

157->211 (diff of 54=2*27)

Aha! Now I see a pattern! The difference of each is 2*number where the number increases by 4 each time.

In a fancy formula, they each differ by 2*(3+4n).

So the 1st term is 1.

The 2nd term is 7, which is 1+2*(3+4(2-2))

The 3rd term is 21, which is 7+2*(3+4(3-2))

The 4th term is 43, which is 21+2*(3+4(4-2))

The 5th term is 73, which is 43+2*(3+4(5-2))

…

And the nth term is the (n-1)th term+2*(3+4(n-2))

That was indeed a tough one!

Thanks Sandra!

I have a quick question about an explicit formula that does not have a common difference. How would I come up with a formula that had a number sequence of 1, 4, 10, 19… for expample. (each number going up by 3, 6, 9, 12..)

You’re starting at 1, so that’s your “seed.”

Then you’re going up by multiples of 3.

Maybe try nth term is (n-1)th term + 3n. Play with that idea.