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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,

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

2 Answers 2

up vote 10 down vote accepted

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.

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

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