Answer:
[tex]e^3[/tex]
Step-by-step explanation:
Given is a series as
[tex]1+\frac{3}{1!} +\frac{3^2}{2!} +...+\frac{3^n}{n!} +...[/tex]
Recall the expansion of
[tex]e^x = 1+x+\frac{x^2}{2!} +...+\frac{x^n}{n!} +...[/tex]
This expansion is valid for all real values of x.
Comparing this with our series we find that x =3
Hence the given series =[tex]e^3[/tex]
Thus we find that the given series can be recognized with the expansion of exponential series with powers of e and here we see that power of e is 3.
So the given Taylor series is equivalent to
[tex]e^3[/tex]