# series and sequences

• Sep 22nd 2012, 02:40 AM
lovelife
series and sequences
Express 1+x^2/(1-x)(1+2x) in partial fraction .
Hence ,findthe constant term in the expansion if 1+x^2/-3x(1-x)(1+2x) in ascending power of x .

I had done the partial fraction ,the answer is 1+x^2/(1-x)(1+2x)= -1/2+2/3(1-x)+5/6(1+2x) .
so can help me solved the next step ?! thanks for help !

• Sep 23rd 2012, 12:01 AM
johnsomeone
Re: series and sequences
Take each of the terms in your partial fraction, and consider it as the sum of a geometric series. Since you're only asked to find the constant term, the only part of the geometric series you'll need will be the leading 1. Thus, assuming your work is right (I didn't check), the constant term when written in powers of x looks to be:
(-1/2)+(2/3)(1) + (5/6)(1) = (-3+4+5)/6 = 1. (Those 1's come from the constant terms in the two geometric series.)

Another way to see this is that the constant term, when expanded in powers of x, must be the value of the function at x=0, hence must be 1.

Another way to see this is to observe that $\displaystyle \frac{x^2}{(1-x)(1+2x)} = \frac{x}{1-x} \frac{x}{1+2x} = \frac{x}{1-x} \frac{(-1/2)(-2x)}{1-(-2x)}$.

Then from the geometric series $\displaystyle \frac{1}{1-t} = 1 + t + t^2 + t^3 + ...$, see that $\displaystyle \frac{t}{1-t} = t + t^2 + t^3 + t^4 + ...$.

Thus $\displaystyle \frac{x^2}{(1-x)(1+2x)} = \frac{x}{1-x} \frac{(-1/2)(-2x)}{1-(-2x)} = \left\{x + x^2 + x^3 +... \right\} \left\{ (-1/2)( (-2x) + (-2x)^2 + (-2x)^3 + ....) \right\}$,

Thus $\displaystyle \frac{x^2}{(1-x)(1+2x)}$ will have no constant term (constant term = 0) when expanded into powers of x, since when those two series are multiplied, there is no constant term.

Or you could note that $\displaystyle \frac{1}{(1-x)(1+2x)} = \frac{1}{1-x} \frac{1}{1-(-2x)} = \left\{1 + x + x^2 + x^3 +... \right\} \left\{ 1 + (-2x) + (-2x)^2 + (-2x)^3 + ....) \right\}$,

which multiplies out to have a constant term 1. So then $\displaystyle \frac{x^2}{(1-x)(1+2x)} = x^2\frac{1}{(1-x)(1+2x)}$ must have $\displaystyle x^2$ and higher power terms only.

Anyway, since $\displaystyle \frac{x^2}{(1-x)(1+2x)}$ has no constant term, the whole expression $\displaystyle \left(1 + \frac{x^2}{(1-x)(1+2x)} \right)$ must have constant term 1.