Madison and Tyler each wrote an expression that is equivalent to The expressions they created are shown in the table.
2 answers:
Answer: Tyler used the distributive property.
Step-by-step explanation:
For the expression:
Madison wrote: which is by commutative property of addition.
- The commutative property of addition says that
, for a,b be any real number
Tyler wrote: which is by distributive propery.
- The distributive property says that
, for a,b,c be any real numbers.
Hence, the last option is correct.
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(a) Suppose 
is a solution for this recurrence, with 
. Then
So we expect a general solution of the form
With
, we get four equations in four unknowns:
So the particular solution to the recurrence is
(b) Let
be the generating function for 
. Multiply both sides of the recurrence by 
and sum over all 
.
From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of, ideally matching the solution found in part (a).
So we expect a general solution of the form
With
So the particular solution to the recurrence is
(b) Let
From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of