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## Anyone recognize this series expansion??

$$1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{51t^4}{4!}+.....$$

I looks kind of like $e^t$ but i am not sure how to deal with it.

Can I factor something... I kind of suck at these. Someone give me a hint.
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 Recognitions: Homework Help Science Advisor I don't quite see the pattern... how do you get from 27 to 51, and what comes next?

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 Quote by nicksauce I don't quite see the pattern... how do you get from 27 to 51, and what comes next?
Oh...yes that should be an 81

$$1+3t+\frac{9t^2}{2!}+\frac{27t^3}{3!}+\frac{81t^4} {4!}+.....$$

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## Anyone recognize this series expansion??

So you have (3t)^0 / 0! + (3t)^1 / 1! + (3t)^2 /2! + (3t)^3 / 3! + ...
Surely you can see what this function is?

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 Quote by nicksauce So you have (3t)^0 / 0! + (3t)^1 / 1! + (3t)^2 /2! + (3t)^3 / 3! + ... Surely you can see what this function is?
I can now! By the way what the Christ is 0! ??? I want to say that for some strange reason it is 1...... but I don't know why??

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 Quote by Saladsamurai I can now! By the way what the Christ is 0! ??? I want to say that for some strange reason it is 1...... but I don't know why??
0!=1, by definition.

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 Quote by cristo 0!=1, by definition.
Yeah. That is what I thought.... though I thought that there was more to it than 'because the math gods said so.'

But I'll take it if that's all there is to it

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