# Anyone recognize this series expansion?

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.

Homework Helper
I don't quite see the pattern... how do you get from 27 to 51, and what comes next?

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

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

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

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??

Staff Emeritus
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.

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