0
$\begingroup$

Determinate the coefficient of $x^k $ of the generating function $f(x)=\frac{x}{1-3x}$

Im trying to do this exercise but something seems wrong. If someone has hints I would appreciate. Thanks in advance So, $f (x)=\frac{x}{1-3x}=- \frac{1}{3} \frac{1}{1- \frac{1}{3x}}=- \frac{1}{3} \sum ( \frac{1}{3x})^k$

So the coefficient is $-\frac{1}{3} ( \frac{1}{3})^k = - \frac{1}{3 \cdot 3^k} $

  • 0
    The $x$ in your expression on the far right is still on the denominator of the fraction, not on the numerator... Note that $\frac{x}{1-3x}=x\cdot \left(\frac{1}{1-3x}\right) = x\cdot \left(\sum\dots\right)$ – 2017-02-22

2 Answers 2

0

$$\frac {1}{1-x}= \sum_{n=0}^{\infty} x^n$$ $$\frac {1}{1-3x}= \sum_{n=0}^{\infty} 3^nx^n$$ $$\frac {x}{1-3x}= \sum_{n=0}^{\infty} 3^nx^{n+1}$$ Now adjust the indeces.

0

Don't divide by $3x$, so that $$\frac{x}{1-3x} = x\frac{1}{1-3x}= x\sum_{n=0}^\infty (3x)^n=\sum_{n=0}^\infty 3^n x^{n+1}$$ and thus the coefficient is $3^{k-1}$