How Do You Calculate the Maclaurin Series for f(x) = 5(x^2)sin(5x)?

[lovelyasha]

Homework Statement

Find the Maclaurin series of the function f(x) = 5(x^2)sin(5x)

Homework Equations

\sum(Cn*x^n)

The Attempt at a Solution

I'm supposed to enter in c3-c7
I already know that c4 and c6 are 0 because the derivative is something*sin(0)=0

but for the odd numbered c's I am having problems...
i know that the taylor series for sinx = \sum((-1^n)*x^(2n+1))/(2n+1)!
so i just substituted in 5x and multiplied by 5x^2 and got
5 \sum ((-1^n)*(5^(2n+1)*x^(2n+3))/(2n+1)!

so for c3 i got 5(-1^3)(5^7)/(7)! = -5^8/7!
but I am not getting the answer right for this. Can someone please explain what I am doing wrong.

 

[Avodyne]

c3 corresponds to n=0 in your series, not n=3 (because you have x^(2n+3), not x^n).

 

[Dick]

Putting n=3 in your formula does not give you c3. It gives you c9. If you want to find ck then the power of x in the formula should be k. The power of x in the formula is 2n+3. So if you want to find c3 set 2n+3=3. You want n=0.

 

[lovelyasha]

Wow, thank you guys! That actually makes sense. I asked my teacher and he wasn't much help, but I guess he didn't know that I was plugging in the wrong n's. I have the right answers now. Thanks again :)