Prove by induction that $n!>2^n$ - Mathematics Stack Exchange
Apply that to the product $$\frac{n!}{2^n}\: =\: \frac{4!}{2^4} \frac{5}2 \frac{6}2 \frac{7}2\: \cdots\:\frac{n}2$$ This is a prototypical example of a proof employing multiplicative telescopy. …
How do you simplify #(n!)/((n-2)!)#? - Socratic
2015年11月18日 · 53533 views around the world You can reuse this answer ...
The idea behind the sum of powers of 2 - Mathematics Stack …
2016年10月29日 · $2^{n+1} - 1 = 2^n + 2^{n-1} + 2^{n-2} . . . + 2^1 + 2^0$ Suppose we take 2^n in the sum. We know since these are powers of two, that the previous term will be half of 2^n, …
Proof by induction: $2^n > n^2$ for all integer $n$ greater than $4$
I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list: Prove, using mathematical induction, that $2^n > n^2$ for all integer n greater …
Formula for $1^2+2^2+3^2+...+n^2$ - Mathematics Stack …
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Comparing $2^n$ to $n!$ - Mathematics Stack Exchange
Comparing $\;2^n\;$ and $\;n!\;$ I need to rewrite one or the other so they can be more easily compared (i.e. have similar form).
Which has a higher order of growth, n! or n^n? [duplicate]
Since n is, by definition, greater than n -1 or n-2, shouldn't any n^n, which is the product of n number of n's, be greater than n!, which is the product of an n number of integers less than n? …
calculus - Find the value of sum (n/2^n) - Mathematics Stack …
2016年4月6日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
calculus - Limit of $n!/n^n$ as $n$ tends to infinity - Mathematics ...
2015年8月8日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …
Exactly how does the equation n (n-1)/2 determine the number of …
2017年4月2日 · $\begingroup$ What $\binom{n}{2}$ counts is the number of ways of selecting two elements from a set without replacement. $\endgroup$ – N. F. Taussig Commented Oct …