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It can often happen in maths that if you miss a definition

or an idea, the rest of the day seems to be in a foreign language.

But do not late strange words like sequence and a(sub n)

bamboozle you, because this is a lot easier than you think.

Here is a simple sequence

1, 3, 5, 7, 9, ... what is the next number ?

Of course you can see it is 11.

Was that really hard ?

9) OK, 1, 1/3, 1/5, 1/7, 1/9, what is the next number ?

Well, I should hope that you see it is 1/11

Right, we have seen the pattern, but could you say what that is in words?

The numbers on the bottom are the odd numbers.

When we write 1/number, that is called the reciprocal.

So, 1, 1/3, 1/5, are the reciprocals of odd numbers.

Yes, yes you knew that, but now we have to put "odd numbers"

into the mysterious and strange language of maths.

This is no time to run out of the room with the screaming had-dabs.

This is the bit where people say I have no idea what comes next,

but I hope to show you, you CAN do this.

First think about even numbers, like {2, 4, 6, 8, ...}

They are all divisible by 2. They are {1, 2, 3, 4, ...}*2

They are all of the form 2n where n = {1, 2, 3, 4, ...}

But what about odd numbers.

It is not enough to just say that they not the even ones.

We have to say that they are ONE MORE than the evens.

Odd numbers are of the form 2n + 1 where n = {1, 2, 3, 4, ...}

Now we are nearly there.

The sequence 1, 1/3, 1/5, 1/7, 1/9, .. that we have called

the reciprocals of odd numbers are all of the form

1/(2n + 1) where n = {1, 2, 3, 4, ...}

Now to that other thing. What on earth is a(sub n) you might wonder?

I am sorry to say it is not an interesting sandwich.

In fact a(sub n) just refers to the GENERAL TERM, or you might say the nth term.

It seems like we have found that already.

The nth term, a(sub n) = 1/(2n + 1) where n = {1, 2, 3, 4, ...}

Did you get all that ? If you are still a bit puzzled, read it through again slowly.

19) a(sub n) = (3 + 5n^2) / (n + n^2)

That is a statement. You forgot to ask the question, but I will

assume you have to work out what the first few terms are.

If that is correct, all you have to do is substitute n = {1, 2, 3, 4, ...}

n = 1: (3 + 5*1^2) / (1 + 1^2) = 4

n = 2: (3 + 5*2^2) / (2 + 2^2) = 23/6

n = 3: (3 + 5*3^2) / (3 + 3^2) = 4

n = 4: (3 + 5*4^2) / (4 + 4^2) = 83/20

I hope that you do understand what is going on now,

and will be able to have a go at these next time.

Regards - Ian

9) The value of n indicates the term in the sequence, like n=2 is 1/3, n=3 is 1/5, which is shown in your problem. The general form for the sequence is:

a(sub n) = 1/(2n-1)

19) I assume you are supposed to come up with some of the numbers of the sequence, as you simply stated the general form for some sequence. For that, simply plug n=1 into the general form to get the first value, n=2 to get the second value, and so on.