# Thread: Recursive formula to a power series

1. ## Recursive formula to a power series

sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

What is the recursive formula for this that connects the nth term with the (n+1)th term?

2. Originally Posted by Sam1111
sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

What is the recursive formula for this that connects the nth term with the (n+1)th term?

$\sum_{k=0}^{\infty} \frac{(-1)^{k+1}x^k}{(k+2)!3^{2k}}$

so:

$a_k=\frac{(-1)^{k+1}x^k}{(k+2)!3^{3k}}$

so:

$a_{k+1}=\frac{(-1)^{k+1}(-1)x^kx}{(k+2)!(k+3)3^{3k}3^2}$

CB

3. Originally Posted by Sam1111
sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

What is the recursive formula for this that connects the nth term with the (n+1)th term?

Let's see: $\sum\limits_{k=0}^\infty \frac{\left(-1\right)^{k+1}x^k}{(k+2)!3^{2k}}$
Copy my formula putting the mouse's pointer on it, and don't forget to open ALWAYS with $$and close with$$