# sigma (c_n * x^n) from 0 to inf

#### s3a

Suppose that

Find the first few coefficients.
?
?
?
?
?

of the power series.
?

I get that R = 6 (which is correct) but I don't know how to find c_0 to c_4. What I think I need to do is isolate x^n and then the rest is the c stuff. So I got sigma from 0 to inf of (-1/6)^n * x^n so (-1/6)^n so, shouldn't that be by C_n values?

This is what I get (which is apparently wrong according to my school's computer system):

C_0 = (-1/6)^0 = 1
C_1 = (-1/6)^1 = -1/6
C_2 = (-1/6)^2 = 1/36
C_3 = (-1/6)^3 = -1/216
C_4 = (-1/6)^4 = 1/1296

Any help would be GREATLY appreciated!

#### dwsmith

MHF Hall of Honor
Suppose that

Find the first few coefficients.
?
?
?
?
?

of the power series.
?

I get that R = 6 (which is correct) but I don't know how to find c_0 to c_4. What I think I need to do is isolate x^n and then the rest is the c stuff. So I got sigma from 0 to inf of (-1/6)^n * x^n so (-1/6)^n so, shouldn't that be by C_n values?

This is what I get (which is apparently wrong according to my school's computer system):

C_0 = (-1/6)^0 = 1
C_1 = (-1/6)^1 = -1/6
C_2 = (-1/6)^2 = 1/36
C_3 = (-1/6)^3 = -1/216
C_4 = (-1/6)^4 = 1/1296

Any help would be GREATLY appreciated!
Is there a differential equation associated with this question?

#### autumn

I would use the geometric series.

$$\displaystyle {3x\over 6+x}=3x\left({1/6\over 1-(-x/6)}\right)$$

$$\displaystyle ={x\over 2}\sum_{n=0}^{\infty}\left({-x\over 6}\right)^n$$

$$\displaystyle ={1\over 2}\sum_{n=0}^{\infty}{(-1)^nx^{n+1}\over 6^n}$$

Last edited: