1. ## Finding the formula

Hi, this question involves finding an explicit formula for the given f(x):

$\displaystyle f(x)=1+2x+x^2+2x^3+x^4+....$ that is, its coefficients are $\displaystyle c_{2n=1}$ and $\displaystyle c_{2n+1}=2$ for all $\displaystyle n\geq0$

However, I only knw how to work from the formla to get the power series but not otherwise. Please help.

The answer is $\displaystyle f(x)=\frac{1+2x}{1-x^2}$

2. Originally Posted by noob mathematician
Hi, this question involves finding an explicit formula for the given f(x):

$\displaystyle f(x)=1+2x+x^2+2x^3+x^4+....$ that is, its coefficients are $\displaystyle c_2n=1$ and $\displaystyle c_2n+1=2$ for all $\displaystyle n\leq0$

However, I only knw how to work from the formla to get the power series but not otherwise. Please help.

The answer is $\displaystyle f(x)=\frac{1+2x}{1-x^2}$
Re-write your series as (i) the even terms and (ii) the odd terms with a factor of $\displaystyle 2x$ taken out

$\displaystyle 1 + x^2 + x^4 + x^6 + \cdots + 2x\left(1 + x^2 + x^4 + x^6 + \cdots \right)$

Then use the geometric power series

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 \cdots.$

3. Hello,

Can't you just split the sum ?

$\displaystyle f(x)=\sum_{k=0}^\infty x^{2k}+2\sum_{k=0}^\infty x^{2k+1}=\sum_{k=0}^\infty x^{2k}+2x\sum_{k=0}^\infty x^{2k}=(1+2x)\sum_{k=0}^\infty x^{2k}$