Let's say I have the following sequence:
25, 50, 75, 100, 125, 150
Each time, 25 is added to the previous value.
Now, let's say I have a cumulative version of this:
25, 75, 150, 250
Can I define this using the Fibonacci sequence?
How do I find the nth term, and more importantly, how do I find how many terms I can fit in until it hits a maximum number?
In the "cumulative" version of your sequence, the differences between successive terms increases linearly: $75 - 25 = 50$, then $150 - 75 = 75$, then $250-150 = 100$ ... each difference is $25$ greater than the last. This means that the sequence can be fitted by a quadratic.
Let the quadratic be $an^2 + bn + c$. It should give the terms in your sequence, using $n=0,1,2,\, ...$
We only have to plug in the first three values of $n$ to get enough information to find the quadratic:
\begin{align} a(0)^2 + b(0) + c &= 25\\\\ a(1)^2 + b(1) + c &= 75\\\\ a(2)^2 + b(2) + c &=150\\ \end{align}
The first equation tells us that $c=25$. The second and third together can be solved to give $a=25/2$ and $b=75/2$.
Therefore, the quadratic we want is $$\boxed{\frac{25}{2}n^2 + \frac{75}{2}n + 25\,}$$
Just to make sure that we didn't mess anything up, use $n=3$ as a check:
$$\frac{25}{2}(3)^2 + \frac{75}{2}(3) + 25$$
This does equal $250$, as it should.
The original sequence is $a_n=25n$. The “cumulative sequence” is $$ b_n=\sum_{k=1}^n a_k=\sum_{k=1}^n 25k=25\sum_{k=1}^n k =25\frac{n(n+1)}{2} $$ Example: for $n=4$, $$ 25\frac{4\cdot 5}{2}=250 $$ See Triangular number for the formula.
Now it’s simple to decide what’s the largest integer for which $b_n\le M$, where $M$ is fixed: just solve $$ 25n^2+25n-2M\le 0 $$ which happens for $$ n\le\frac{\sqrt{625+200M}-25}{50} $$ solving a simple quadratic inequality.
and more importantly, how do I find how many terms I can fit in until it hits a maximum number
The sum of the the sequence is
$S_n=k\cdot \frac{n\cdot (n+1)}{2}\quad |\cdot 2\quad |:k$
$n$ is the number of the terms and $k\in \mathbb N$ is an arbitrary constant.
$\frac{2S_n}k=n\cdot (n+1)$
$\frac{2S_n}k=n^2+n$
Completing the square
$\frac{2S_n}k=(n^2+n+\frac14)+\frac14 \quad |-\frac14$
$\frac{2S_n}k-\frac14=(n+0.5)^2 \quad |\sqrt{()}$
We take the positive root only since $n$ is positive.
$\sqrt{\frac{2S_n}k+\frac14}=n+0.5 \quad |-0.5$
$$\boxed{-0.5+\sqrt{\frac{2S_n}k+\frac14}=n}$$
You can say that $S_n$ is the maximum number. If $k=25$ and $S_n=1000$ then you can calculate how many terms you can sum up.
$n\leq -0.5+\sqrt{\frac{2000}{25}+\frac14}=8.45...$
Since n is a whole number you can sum up 8 terms:
$25+50+75+100+125+150+175+200=25\cdot \frac{8\cdot 9}{2}=900$
If you would add the next term ($225$) the sum of the sequence would be greater than $1000$.
