Simplify K!/(K+1)!
Simplify: $\frac{K!}{(K+1)! }$
I know that this simplifies to $\frac{1}{k+1}$, but I don't know how my professor does it. How would you go about solving this with the knowledge required in introductory calculus class? Thanks.
$\endgroup$4 Answers
$\begingroup$$(K+1)!=(K+1)K!$
Now cancel the $K!$s.
$\endgroup$ 0 $\begingroup$$$\frac{K!}{(K+1)!} = \frac{1\cdot 2 \cdot \cdots \cdot K}{1 \cdot 2\cdot \cdots \cdot K \cdot (K+1)} = \frac{1}{K+1}$$
$\endgroup$ $\begingroup$Pretend K=3 That means (K+1)= 4
This means you'd be dividing 3*2*1 by 4*3*2*1. Consider how you'd cancel out multiples by dividing them. Like how (2(5+x))/2 would just equal 5+x.Following that idea we'd pretty much be able to cancel out every number in the numerator, so long as its also in the denominator. This would end up canceling every number except for 4 which equals (K+1). Essentially 1/(K+1) is the answer because 1/(K+1) is all that is left after everything gets canceled.
$\endgroup$ $\begingroup$$$k!=$$ $$k * (k-1) * (k-2) * (k-3)....(4) * (3) * (2) * (1)$$
$$(k+1)!=$$ $$(k+1)[(k) * (k-1) * (k-2) * (k-3)....(4) * (3) * (2) * (1)]=$$ $$(k+1)(k!)$$
Now,
$$\frac{k!}{(k+1)!}=$$
$$\frac{k!}{(k+1) (k!) }=$$
$$\frac{1}{k+1}$$
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