1. ## Basic Porblem

Basic but very annoying question

the series 5, 9, 14, 20 ect
what is tye formular for the nth term

2. Any ideas on this one? Two hints, the next number in the sequence is 27, and it might be worthwhile asking what is the difference between the first and second term, then the second and third term (etc.) is.

Let us know how you go.

3. ## Still probomatic

thanks

the diffrence between t1 and t2 is 4
and t2 and t3 is 5 ect, increaing in increments of 1
i understand that but for some rason i cant see to find the equationn for the nth term,
it was a question on my sisters test and i just cant get it
any ideas?

4. Right, so as n increases, we add more on. This is our first hint, that the n'th term will depend on n somehow.

Let's look at $t_2$, you pointed out that it's 4 more than the first term, that is $t_2 = t_1 + 4.$ Now I want to work n into it somehow, since we noticed it depends on n. So maybe something like $t_2 = t_1 + 1 + 3.$

Hows that?

5. Is this the equation you're thinking of?

where a1 = initial value, r = common ratio for a geometric series.

6. Originally Posted by Pom
Basic but very annoying question

the series 5, 9, 14, 20 ect
what is tye formular for the nth term
First diferences are 4, 5, 6, the second differences are 1, 1, 1, so this is a
quadratic in $n$, so we put:

$
a_n = u n^2 + v n + w
$

$a_1=5$, so $u+v+w=5$
$a_2=9$, so $4u+2v+w=9$
$a_3=14$, so $9u+3v+w=14$

Now solve for $u,\ v,\ \mbox{ and } w$ to get the answer.

Now I make the solution $u=1/2,\ v=5/2\ w=2$, so:

$
a_n=\frac{n^2+5n+4}{2}=\frac{(n+4)(n+1)}{2}
$

RonL