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Simplification Difficulties

  • Thread starter roam
  • Start date
  • #1
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Homework Statement



I want to simplify the following (these are to do with the ratio test in series):

a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]


The Attempt at a Solution



a) [tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

[tex]\frac{(n+1)!}{(2n+1)} \times \frac{(2n-1)}{n!}[/tex]

So, we devide up the factorial terms: [tex]\frac{n \times n+1}{n} = n+1[/tex]

and deviding the rest:

[tex]\frac{2n-1}{2n+1}=-1[/tex]

Therefore the whole thing should simplify to: -(n+1) But the book says it must be [tex]\frac{n+1}{2n+1}[/tex], what's wrong?

b) [tex]\frac{2^{2n+1}}{(2n+1)!} / \frac{2^{2n-1}}{(2n-1)!}[/tex]

[tex]\frac{2^{2n+1}}{(2n+1)!} \times \frac{2^{2n-1}}{(2n-1)!}[/tex]

[tex]\frac{4n . 2}{4n . 1/2} = 4[/tex]

(2n-1)!/(2n+1)! = [tex]\frac{(2n)(2n-1)}{(2n)(2n+1)}[/tex]

So based on what I've done my answer has to be -4 but the real answer has to be:
4/(2n+1)(2n)

Can anyone please help me with this simplifications...
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
N!=N(N-1)(N-2)(N-3)!

so what is (n+1)!=? (Replace N by n+1)
 
  • #3
statdad
Homework Helper
1,495
35
Note that

[tex]
\frac{2n-1}{2n+1}=-1[/tex]

is not correct - you cannot cancel the pair of [tex] 2n [/tex] terms from this. You are
correct in saying that

[tex]
\frac{(n+1)!}{n!} = n+1
[/tex]

In your second problem, this first step is not correct. That is, you wrote

[tex]
\frac{2^{2n+1}}{(2n+1)!} \div \frac{2^{2n-1}}{(2n-1)!} = \frac{2^{2n+1}}{(2n+1)!} \times \frac{2^{2n-1}}{(2n-1)!}
[/tex]

and this is not correct.
 
  • #4
1,266
11
Note that

[tex]
\frac{2n-1}{2n+1}=-1[/tex]

is not correct - you cannot cancel the pair of [tex] 2n [/tex] terms from this.
Why? So, what do we have to do with it???

P.S. It was just a typo in my second problem... I figured it out anyway. :smile:


N!=N(N-1)(N-2)(N-3)!

so what is (n+1)!=? (Replace N by n+1)
I understood, thanks!
 
  • #5
statdad
Homework Helper
1,495
35
"Why? So, what do we have to do with it???"

I'm not sure exactly what you mean - if what you've typed is correct, you can't simplify

[tex]
\frac{2n-1}{2n+1}
[/tex]

and get [tex] -1 [/tex].
 
  • #6
1,266
11
If it's wrong then how do you simplify (2n-1)/(2n+1)?

Yes, what I've typed is correct:

[tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]

And the correct answer I should get is:

[tex]\frac{n+1}{2n+1}[/tex]

If I devide (n+1)!/n! = n+1 which goes on the numerator, what should I do with (2n-1)/(2n+1)?
 
  • #7
statdad
Homework Helper
1,495
35
If it's wrong then how do you simplify (2n-1)/(2n+1)?

There isn't much you can do - you can't cancel terms that are added, only common factors from a product. The only thing that comes to my mind now is this:

[tex]
\frac{2n-1}{2n+1} = \frac{(2n+1)-2}{2n+1} = 1 - \frac{2}{2n+1}
[/tex]

and that provides you with absolutely no help. This is why I'm confused - I don't know how you are to simplify the expression you are given and obtain your stated answer.

But again,

[tex]
\frac{2n-1}{2n+1} \ne -1
[/tex]

Think this way: if the left hand side truly did simplify to [tex] -1 [/tex], it would have
that value for every possible value of [tex] n [/tex] (because it's a constant). But, when
[tex] n = 10 [/tex], the left hand side is 19/21.

Perhaps there is a typographical error in your text.
 
  • #8
1,266
11
Perhaps there is a typographical error in your text.
Are you sure it's an error??

here's the actual problem from my textbook:

http://img23.imageshack.us/img23/635/ddfy.jpg [Broken]
 
Last edited by a moderator:
  • #9
52
0
Wait, did you simplify this:
[tex]\frac{s_{n+1}}{s_{n}} = \frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)}/\frac{n!}{1 \times 3 \times 5 \times ... \times (2n-1)}[/tex]

to get what you wrote originally in part a), which was this:
[tex]\frac{(n+1)!}{(2n+1)}/\frac{n!}{2n-1}[/tex]?

If so, that wasn't simplified correctly. First rewrite as a multiplication:
[tex]\frac{(n+1)!}{1 \times 3 \times 5 \times ... \times (2n+1)} \times \frac{1 \times 3 \times 5 \times ... \times (2n-1)}{n!}[/tex]

1*3*5*...*(2n+1) is the same as 1*3*5*...*(2n-1)*(2n+1), so if you look at the denominator of the 1st fraction and the numerator of the 2nd fraction, everything cancels out except for the 2n+1:
[tex]\frac{(n+1)!}{2n+1} \times \frac{1}{n!}[/tex]

You can figure out the rest. (I hope that I didn't give away too much.)


01
 
  • #10
1,266
11
1*3*5*...*(2n+1) is the same as 1*3*5*...*(2n-1)*(2n+1)
Actually I think you meant that it is the same as 1*3*5*...*(2n-1)*(2n)*(2n+1) :biggrin:

That makes sense now, thank you.

EDIT:

I have another question (which is very similar to the one above):

[tex]\frac{(n+1)!}{3\times5 \times7\times...(2n+3)} \times \frac{3\times5\times7\times... \times (2n+1)}{n!}[/tex]

Does it equal to:

[tex]\frac{n+1}{(2n+2)(2n+3)}[/tex]

Since 3*5*7*...*(2n+3) is the same as 3*5*7*...*(2n+1)*(2n+2)*(2n+3)

Is this right? :rolleyes:
 
Last edited:
  • #11
52
0
Actually I think you meant that it is the same as 1*3*5*...*(2n-1)*(2n)*(2n+1) :biggrin:
Nope, I meant that 1*3*5*...*(2n+1) = 1*3*5*...*(2n-1)*(2n+1). We are multiplying a series of odd numbers here. 2n-1 is odd, 2n+1 is odd, but 2n is even.

[tex]\frac{(n+1)!}{3\times5 \times7\times...(2n+3)} \times \frac{3\times5\times7\times... \times (2n+1)}{n!}[/tex]

Does it equal to:

[tex]\frac{n+1}{(2n+2)(2n+3)}[/tex]

Since 3*5*7*...*(2n+3) is the same as 3*5*7*...*(2n+1)*(2n+2)*(2n+3)

Is this right? :rolleyes:
No. 3*5*7*...*(2n+3) = 3*5*7*...*(2n+1)*(2n+3). 2n+2 shouldn't be there -- it's also even.


01
 
  • #12
1,266
11
Oh!! I didn't notice that. So it simplifies to n+1/2n+3. Thanks for your post, it was really helpful!
 

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