Video Transcript
Write down a recursive formula for
the sequence three, nine, 21, 45, 93, and so on.
Now, again, good starting point is
to actually say well what’s the difference between each term and the next term. So, to get from three to nine, I
need to add six. But to get from nine to 21, I need
to add 12. To get from 21 to 45, I need to add
24. And to get from 45 to 93, I need to
add 48. So, this isn’t an arithmetic
sequence. This is a bit of a trickier
prospect.
Now, if you look at each term, and
then you look at the differences, we can see that in the first case we start off
with three but we’re adding six. Then we start, then we start off
with nine, and we’re adding 12. We start off with 21 and we’re
adding 24. We start off with 45, we’re adding
48. These differences are always three
more than the previous term. Now, if we label our terms 𝑎 one,
𝑎 two, 𝑎 three, 𝑎 four, 𝑎 five, and so on, the second term is equal to the first
term plus six. And the third term is equal to the
second term plus 12. But we said that the difference is
three bigger than the term- the previous term itself. So, this difference is three bigger
than this.
So, six is 𝑎 one plus three. And for working out a third term,
that 12, that difference there, is the second term plus three again. So, the third term is equal to the
second term plus the second term plus three. And that’s gonna be the case
generally. The 𝑛 plus oneth term is equal to
the 𝑛th term plus the 𝑛th term plus three. And given that this is just three
values added together — I don’t need those parentheses — that gives me 𝑎 𝑛 plus 𝑎
𝑛 plus three. That’s two lots of 𝑎 𝑛 plus
three. So, 𝑎 𝑛 plus one is equal to two
𝑎 𝑛 plus three.
And now, we need to think about the
starting conditions. Well, the first term 𝑎 one is
equal to three. And to generate terms 𝑎 one, 𝑎
two, 𝑎 three, 𝑎 four, and so on, I need to put 𝑛 equal to one, two, three, and so
on. So, that’s it; there’s my
formula. Now, I could adjust this slightly
as well like I did the last time. And that will give me the 𝑛th term
is equal to two times the 𝑛 minus oneth term plus three. And again, I’d have to adjust the
starting point of 𝑛, 𝑛 is greater than or equal two, so that I don’t end up with
the term 𝑎 zero.
