Finding the formula

**noob mathematician** · November 7th 2009, 06:18 AM

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

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

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

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

**Danny** · November 7th 2009, 06:23 AM

Originally Posted by noob mathematician

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

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

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

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

Re-write your series as (i) the even terms and (ii) the odd terms with a factor of $2x$ taken out

$<br /> 1 + x^2 + x^4 + x^6 + \cdots + 2x\left(1 + x^2 + x^4 + x^6 + \cdots \right)<br />$

Then use the geometric power series

$<br /> \frac{1}{1-x} = 1 + x + x^2 + x^3 \cdots.<br />$

**Moo** · November 7th 2009, 06:24 AM

Hello,

Can't you just split the sum ?

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

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