My Math Forum Relating Pascal's Triangle and Compound Interest

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 February 5th, 2016, 11:04 AM #1 Newbie   Joined: Feb 2016 From: Oregon Posts: 2 Thanks: 0 Relating Pascal's Triangle and Compound Interest I hope that this is the right forum to post this question. In exploring the relationship between compound interest and Pascal's Triangle, I've become stuck at trying to explain why $\frac{(n-1)^{n-2}}{n!^2}= \frac{1}{n^n}$ I would greatly appreciate it if someone could explain the intermediary steps taken to come to this conclusion.
February 5th, 2016, 01:21 PM   #2
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 Originally Posted by Blackwatercity I hope that this is the right forum to post this question. In exploring the relationship between compound interest and Pascal's Triangle, I've become stuck at trying to explain why $\frac{(n-1)^{n-2}}{n!^2}= \frac{1}{n^n}$ I would greatly appreciate it if someone could explain the intermediary steps taken to come to this conclusion.
It's not true! You must be missing something.

 February 5th, 2016, 02:32 PM #3 Newbie   Joined: Feb 2016 From: Oregon Posts: 2 Thanks: 0 Apparently what I had was completely wrong, so could somebody tell me if the equations below are correct? At least one of them should be incorrect. Thanks First: $\frac{\prod_{k=0}^{n-1}(n-1)!}{\prod_{k=0}^{n}n!}= \frac{(n-1)!^n}{n!^{n+1}}$ Second: $\frac{\prod_{k=0}^{n}(n-k)!}{\prod_{k=0}^{n-1}(n-1-k)!}= (n-1)!$ Third: $\frac{\prod_{k=0}^{n}k!}{\prod_{k=0}^{n-1}k!}= (n-1)!$
 February 5th, 2016, 03:00 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 The first one is right, the second and third ones are wrong. They equal $n!$. Thanks from Blackwatercity

pascal triangle for compound interest

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