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Mathematics 13 Online
OpenStudy (anonymous):

Write the following power series as a function e) \(\sum_{n=0}^{\infty}\frac{2^{n}x^{3n}}{n!}\)

OpenStudy (kinggeorge):

Just for easier reading: \[\Large \sum_{n=0}^{\infty}\frac{2^{n}x^{3n}}{n!}\]

OpenStudy (anonymous):

ok..

OpenStudy (kinggeorge):

This isn't actually my best area of expertise unfortunately. Perhaps @joemath314159 help out with this?

OpenStudy (anonymous):

ok any help would be apreciated

OpenStudy (anonymous):

isnt it:\[\frac{2^nx^{3n}}{n!}=\frac{(2x^3)^n}{n!}\]So that summation is really just\[e^{2x^3}\]?

OpenStudy (anonymous):

mm, that i really dont know..maybe George can answer..

OpenStudy (kinggeorge):

I think that's right.

OpenStudy (anonymous):

Im using the fact that:\[e^x=\sum_{n=0}^\infty \frac{x^n}{n!}\]in place of x we have 2x^3 though.

OpenStudy (anonymous):

ok is this end solution, or is there someting needed to be add?

OpenStudy (anonymous):

That should be it. they wanted you to write that power series as a function, and it turned out to be e^(2x^3). Done and done :)

OpenStudy (anonymous):

ok thank you very much both of you :) i am happy that i found this website and you guys

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