binomial theorem / expansion

mattking32

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I have been stuck at a question for 3 hours, so I decided to look up the answer on the internet. This is the answer:Screen Shot 2018-12-10 at 11.08.27 PM.jpg
 
Need help with binomial theorem

I have been stuck in this working out for 2 hours. Can anyone explain ( working out would be great ) how did he get 2(n-2)! = 4 * 6(n-3)! ?

Screen Shot 2018-12-10 at 11.08.27 PM.jpg
 
I have been stuck in this working out for 2 hours. Can anyone explain ( working out would be great ) how did he get 2(n-2)! = 4 * 6(n-3)! ?

View attachment 10634
attachment.php

expand

3! = 6
2! = 2

(n-2)! = (n-2)*[(n-3)!]

and simplify
 
Cross multiplying, that is if \(\displaystyle \frac{a}{b} = \frac {c}{d}\) then \(\displaystyle a*d = b*c\) and the n! cancels
Hey, that is what I said!!

Not a big deal, as reading it twice the OP might remember it!
 
I have been stuck in this working out for 2 hours. Can anyone explain ( working out would be great ) how did he get 2(n-2)! = 4 * 6(n-3)! ?

View attachment 10634
Well, I am lost.

We start with a summation symbol referencing k, but nothing to sum and neither a starting nor ending value for k. We have a coefficient of zero to the n? Huh? Is this some sort of notation that I am unfamiliar with?

In any case, how do we possibly "know" from the binomial theorem that

\(\displaystyle \dbinom{n}{3} = 4 * \dbinom{n}{3}.\)

\(\displaystyle \dbinom{n}{3} = 4 * \dbinom{n}{3} \implies 4 = 1 \text { or } \dbinom{n}{3} = 0.\)

Nor is it generally true that

\(\displaystyle \dfrac{n!}{3! * (n - 3)!} = 4 * \dfrac{n!}{2! * (n - 2)!}\)

For example,

\(\displaystyle \dfrac{6!}{3! * 3!} = \dfrac{6 * 5 * 4}{3 * 2} = 20.\) But

\(\displaystyle 4 * \dfrac{6!}{2! * 4!} = 4 * \dfrac{6 * 5}{2} = 4 * 15 = 60.\)

I agree of course that

\(\displaystyle \dbinom{14}{3} = \dfrac{14!}{3! * 11!} = \dfrac{14 * 13 * 12}{3 * 2} =4 * \dfrac{14 * 13 * 3}{3 * 2} =\)

\(\displaystyle 4 * \dfrac{14 * 13}{2} = 4 * \dfrac{14 * 13 * 12!}{2 * 12!} = 4 \dfrac{14!}{2! * 12!} = 4 * \dbinom{14}{2}.\)

Bizarre.
 
I have been stuck in this working out for 2 hours. Can anyone explain ( working out would be great ) how did he get 2(n-2)! = 4 * 6(n-3)! ?

View attachment 10634

I think your direct question has been answered; are you satisfied with the idea of cross-multiplication?

Now we are starting to get confused by the typos and lack of complete statement of the problem.

I think the sigma is supposed to have k=0 underneath and n at the top; presumably some braces were omitted from the LaTeX. Did you do this, or get it from elsewhere?

I think the statement that \(\displaystyle \dbinom{n}{3} = 4\dbinom{n}{3}\) is a typo for \(\displaystyle \dbinom{n}{3} = 4\dbinom{n}{2}\), judging by what follows, and is motivated by something in the actual problem, which you have not stated.

Since you didn't ask about those things, presumably they are not a problem for you; but it would be kind to us if you had (as we ask) stated the whole problem you are working on, or else only copied the part you are asking about ...
 
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