3
$\begingroup$

I'm re-learning factorials, and I encountered this exercice, but the solution had a diferent result than I got, and no matter how much I try to search, I can't find an explanation to the last step of this:

$$\frac{100!+98!}{100!-98!}\iff\frac{(100\times 99\times 98!)+98!}{(100\times 99\times 98!)-98!}\iff\frac{98!(9900+1)}{98!(9900-1)}\iff\frac{9901}{9899}$$

I understand $100 \times 99 = 9900$ but where does the $1$ come from? and where does the $98!$ go?

Can someone please explain me that last step?

When I calculated myself I simply canceled and got:

$$\frac{100 \times 99 \times 98! + 98!}{100 \times 99 \times 98! - 98!} = \frac{98!}{98!} = 1$$

Where am I going wrong? Thanks,

$\endgroup$
  • 3
    $\begingroup$ The $1$ comes from the fact that $98! \times 1 = 98!$, hence $100 \times 99 \times 98! + 1 \times 98! = 98! \times \left( 100 \times 99 + 1 \right) = 98! \times 9901$. $\endgroup$ – Sasha Jan 24 '12 at 1:31
  • 1
    $\begingroup$ Apply the rule ab+a = a(b+1) in the numerator with a=100x99 and b=98!. $\endgroup$ – wnvl Jan 24 '12 at 1:34
10
$\begingroup$

It's the distributive law: $ac+bc=(a+b)c$ with $a=100\times 99$, $b=1$, and $c=98!$. The $b$ is invisible in $100\times99\times98!+98!$ because multiplication by $1$ does nothing.

By the way, you should be using "$=$" instead of "$\Leftrightarrow$". The arrows belong between claims or equations that can be true or false, but here you have expressions that stand for numerical values. Such things take equals signs.

$\endgroup$
3
$\begingroup$

$100 \times 99 \times 98! + 98! = (100 \times 99 + 1 ) \times 98! = (9900 + 1)\times 98!$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.