Given f(r)=r(r−1)!\mathrm{f}(r) = \frac{r}{(r-1)!}, show that f(r−1)−f(r)=r2−3r+1(r−1)!\mathrm{f}(r-1) - \mathrm{f}(r) = \frac{r^2 - 3r + 1}{(r-1)!}.

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Worked solution

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1. Express $\mathrm{f}(r-1)$ and set up the expression for the difference.

Substitute r−1r-1 into f\mathrm{f}:
f(r−1)=r−1(r−2)!\mathrm{f}(r-1) = \frac{r-1}{(r-2)!}

Set up the difference:
f(r−1)−f(r)=r−1(r−2)!−r(r−1)!\mathrm{f}(r-1) - \mathrm{f}(r) = \frac{r-1}{(r-2)!} - \frac{r}{(r-1)!}

2. Find a common denominator using the property of factorials and simplify.

Recognise that (r−1)!=(r−1)(r−2)!(r-1)! = (r-1)(r-2)!. Multiply the numerator and denominator of the first term by (r−1)(r-1):
=(r−1)(r−1)(r−1)!−r(r−1)!= \frac{(r-1)(r-1)}{(r-1)!} - \frac{r}{(r-1)!}
=r2−2r+1(r−1)!−r(r−1)!= \frac{r^2 - 2r + 1}{(r-1)!} - \frac{r}{(r-1)!}
=r2−2r+1−r(r−1)!= \frac{r^2 - 2r + 1 - r}{(r-1)!}
=r2−3r+1(r−1)!(Shown)= \frac{r^2 - 3r + 1}{(r-1)!} \quad \text{(Shown)}

Answer

r2−3r+1(r−1)!\frac{r^2 - 3r + 1}{(r-1)!} (Shown)

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