Answer:
[tex]\begin{aligned}\bullet\ &f(1)=800;f(n)=f(n-1)+900, \text{for $n\ge 2$}\\ \bullet\ & f(n)=900n-100\end{aligned}[/tex]
Step-by-step explanation:
See attachment for the figure.
Using arithmetic sequence with a first term of 800 and a common difference of 900. The general form for such a sequence is given by,
an = a1 +d(n -1)
an = 800 +900(n -1) = 900n -100
If n is the function, this can be written as,
f(n) = 900n -100
When considered as a recursive relation, we find the first term is still 800:
f(1) = 800
and that each term is 900 more than the previous one:
f(n) = f(n-1) +900 . . . . for n ≥ 2
You need to consider that huge numbers of the different answer decisions are debasements of either of these structures, so you should look at them cautiously.
