$$f_{n+1}\geq r^{n-2}+r^{n-3}$$ Factoring out a common term of $r^{n-3}$ from line (2), we get: $$f_{n+1}\geq r^{n-3}(r+1)$$

I don't understand how to factor out r^(n-3) to get that result.

  • $\begingroup$ $r^{n-3} r^{1} = r^{n-3+1} = r^{n-2}$. $\endgroup$ – AnonSubmitter85 Feb 11 '14 at 23:32
  • $\begingroup$ What two variables? $\endgroup$ – Martín-Blas Pérez Pinilla Feb 11 '14 at 23:33
  • $\begingroup$ Which result? btw, r^(n-3) = (1 + (r-1))^(n-3), use extended binomial theorem here $\endgroup$ – abstractnature Feb 11 '14 at 23:33
  • $\begingroup$ @Martín-BlasPérezPinilla r and n $\endgroup$ – compguy24 Feb 11 '14 at 23:34
  • 1
    $\begingroup$ $r^{n-2}+r^{n-3}$ is a polynomial of degree $n-2$ in one variable. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 11 '14 at 23:35

$r$ to any power is a common factor of both. I'm sure you are familiar with the following rule for exponents: $r^a \cdot r^b = r^{a+b}$. The other thing you need to know is that $r^0 = 1$. Using this, we can factor out $r^c$ from any expression of the form $r^x + r^y$:

$$ r^x + r^y = r^c ( r^{x-c} + r^{y-c}) $$

In this case, we have $x=n-2$, $y=n-3$, and $c=n-3$, which gives $r^{n-3}(r+r^0) = r^{n-3}(r+1)$.

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