Question

Select ALL the correct answers.

Answer 1

Answer:

$\begin{aligned}\bullet\ &f(1)=800;f(n)=f(n-1)+900, \text{for n\ge 2 }\\ \bullet\ & f(n)=900n-100\end{aligned}$

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.