Basic but very annoying question
the series 5, 9, 14, 20 ect
what is tye formular for the nth term
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?
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 $\displaystyle t_2$, you pointed out that it's 4 more than the first term, that is $\displaystyle t_2 = t_1 + 4.$ Now I want to work n into it somehow, since we noticed it depends on n. So maybe something like $\displaystyle t_2 = t_1 + 1 + 3.$
Hows that?
First diferences are 4, 5, 6, the second differences are 1, 1, 1, so this is a
quadratic in $\displaystyle n$, so we put:
$\displaystyle
a_n = u n^2 + v n + w
$
$\displaystyle a_1=5$, so $\displaystyle u+v+w=5$
$\displaystyle a_2=9$, so $\displaystyle 4u+2v+w=9$
$\displaystyle a_3=14$, so $\displaystyle 9u+3v+w=14$
Now solve for $\displaystyle u,\ v,\ \mbox{ and } w$ to get the answer.
Now I make the solution $\displaystyle u=1/2,\ v=5/2\ w=2$, so:
$\displaystyle
a_n=\frac{n^2+5n+4}{2}=\frac{(n+4)(n+1)}{2}
$
RonL