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Check the forward differences of the sequence.
If
, then let
be the sequence of first-order differences of
. That is, for n ≥ 1,

so that
.
Let
be the sequence of differences of
,

and we see that this is a constant sequence,
. In other words,
is an arithmetic sequence with common difference between terms of 2. That is,

and we can solve for
in terms of
:



and so on down to

We solve for
in the same way.

Then



and so on down to


Answer:
A. (0,1) and (1,2)
Step-by-step explanation:
I graphed it
It is B because I had the question before in my school
Since you gave no parenthesis: Remember to follow PEMDAS. Note that anything with the power of 0, the answer will become 1
10^0 = 1
8 x 1 = 8
8 is your answer
hope this helps