Final answer:
The sequence generated by the exponential generating function f(x)=3e⁹ˣ is obtained by expanding it into a Taylor series. This results in the sequence starting with 27, followed by 243x, 2,187x²/2, 19,683x³/6, and continuing with the respective exponential terms multiplied by 27 and divided by the factorial of the term's power.
Step-by-step explanation:
To determine the sequence generated by the exponential generating function f(x)=3e³ˣ, we need to understand how to work with exponentials. When cubing a term like 3eˣ, you cube the digit term, which is 3 in this case, in the usual way to get 27, and then multiply the exponent of the exponential term by 3. So eˣ becomes e⁹ (since 3 times 3 is 9). Therefore, f(x) can be written as 27e⁹ˣ. To visualize the sequence, we need to expand f(x) as a Taylor series.
The Taylor series expansion of the exponential function ex is 1 + x/1! + x²/2! + x³/3! + ... and so on. Thus, for e⁹ˣ, the series would be 27(1 + 9x/1! + 81x²/2! + 729x³/3! + ...). This means the series generated by f(x)=3e³ˣ starts with 27, 27×9 = 243x, 27×81/2! = 2,187x²/2, 27×729/3! = 19,683x³/6, and so on. Each term of the sequence can be found by plugging in integer values of x.