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Binomial theorem to prove

  • Thread starter rohan03
  • Start date
1. The problem statement, all variables and given/known data

(1+n)^n≥ 5/2* n^n- 1/2* n^(n-1) for n≥2

2. Relevant equations
i know I have to use this formula
(1+x)^n=1+nx/1!+(n(n-1) x^2)/2!+⋯

3. The attempt at a solution

And you take x=n from my original inequality but after that I have no clue
(1+n)^n=1+n/1! n+(n(n-1) n^2)/2!+⋯
but it seems very complicated !!
 

LCKurtz

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1. The problem statement, all variables and given/known data

(1+n)^n≥ 5/2* n^n- 1/2* n^(n-1) for n≥2

2. Relevant equations
i know I have to use this formula
(1+x)^n=1+nx/1!+(n(n-1) x^2)/2!+⋯

3. The attempt at a solution

And you take x=n from my original inequality but after that I have no clue
(1+n)^n=1+n/1! n+(n(n-1) n^2)/2!+⋯
but it seems very complicated !!
Since n is a positive integer, that binomial expansion has finitely many terms. And since ##n\ge 2## the expansion has at least 3 terms. Try keeping just the last three terms (which you didn't write down) because they are the ones with the higher powers of n.
 
Will do that and write my results here. What I don't know is how to get last terms power to n-1 ?
 
I'll be doing this tomorrow so please check ur post.
 
(1+n)^n=1+n/1 n+(n(n-1) n^2)/2!+(n(n-1) 〖(n-2)n〗^3)/3!…
 
Last edited:

LCKurtz

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(1+n)^n=1+n/1 n+(n(n-1) n^2)/2!+(n(n-1) 〖(n-2)n〗^3)/3!…
Do you know the binomial expansion for ##(a+b)^n##? It doesn't end with a "...". You need to look at the last three terms, not the first terms.
 
if I use just the last three term I do get the result- so is it ok for me to just use last three terms ignoring first few terms- if yes than I have resolved so simple problem- which looked very complicated in the first place- Thank you
 

LCKurtz

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if I use just the last three term I do get the result- so is it ok for me to just use last three terms ignoring first few terms- if yes than I have resolved so simple problem- which looked very complicated in the first place- Thank you
Yes. That is why there is a ##\ge## sign in the statement of the problem. Leaving out the other terms makes it smaller.
 
thank you- this forum is blessing to someone relatively new to pure maths.
 
I'm confused.

Which binomial expansion should I be using for this question and why?

Is it binomial expansion for (a+b)^n or (1+x)^n
 
what do you mean last 3 terms? if its infinite it wouldn't have any last term?
 

LCKurtz

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what do you mean last 3 terms? if its infinite it wouldn't have any last term?
##n## is a positive integer so the binomial expansion is finite:$$
(1+n)^n =\sum_{k=0}^n \binom n k 1^{n-k} n^k$$The last three terms are for ##k=n,n-1,n-2##.
 
Thanks for that. I'm having a blonde moment though when working out the coefficients. I know you use n!/k!(n-k)! However I am little confused when I input for say k = n-1 I get n!/(n-1)!(n-n-1)! How do I expand on this? Many thanks.
 
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Thanks for that. I'm having a blonde moment though when working out the coefficients. I know you use n!/k!(n-k)! However I am little confused when I input for say k = n-1 I get n!/(n-1)!(n-n-1)! How do I expand on this? Many thanks.
A factorial for natural numbers is the product of all consecutive integers from 1 upto n. That is,

[tex]n! = n\cdot (n-1) \cdot (n-2) ........ 2\cdot 1[/tex]

Use this definition in both the numerator and denominator of the expression. What do you get?



I'm confused.

Which binomial expansion should I be using for this question and why?

Is it binomial expansion for (a+b)^n or (1+x)^n
https://www.physicsforums.com/library.php?do=view_item&itemid=869
 
so say i was doing k = n-1

n(n-1)/n(n-1)(0)! = n(n-1)/0! = n(n-1)? have I dont that correct??
 

LCKurtz

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so say i was doing k = n-1

n(n-1)/n(n-1)(0)! = n(n-1)/0! = n(n-1)? have I dont that correct??
No.$$
\binom n {n-1}=\frac {n!}{(n-1)!(n-(n-1))!}=\frac {n!}{(n-1)!(n-n+1)!}
=n$$
 
thank you for your help so far.

I wanna see if i have expanded this right then. Definetly need more practice on this.

(1+n)^n = 1^n + 1^n + n^n-1 + 1/2n^n-2 + [itex]\cdots[/itex]
 
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thank you for your help so far.

I wanna see if i have expanded this right then. Definetly need more practice on this.

(1+n)^n = 1^n + 1^n + n^n-1 + 1/2n^n-2 + [itex]\cdots[/itex]
Use parenthesis!! I can't figure out the equation completely, but it is surely incorrect as you won't get 1n twice. The binomial expansion is given as,

[tex](a+b)^n = \binom{n}{0} a^{n}b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + .......[/tex]

Keeping in mind what LCKurtz said, what do you get if you put a=1, b=n, and n=n in the above equation?

Also, did you check out the link in my previous post??
 
[itex]\frac{n!}{0!(n-0)!}[/itex]1[itex]^{n}[/itex] + [itex]\frac{n!}{(n!)(n-n)!}[/itex]n[itex]^{n}[/itex] + [itex]\frac{n!}{(n-1)!(n-(n-1))}[/itex]n[itex]^{n-1}[/itex]+[itex]\frac{n!}{(n-2)!(n-(n-2))}[/itex]n[itex]^{n-2}[/itex] + [itex]\cdots[/itex]

= 1 + n[itex]^{n}[/itex] + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?
 
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[itex]\frac{n!}{0!(n-0)!}[/itex]1[itex]^{n}[/itex] + [itex]\frac{n!}{(n!)(n-n)!}[/itex]n[itex]^{n}[/itex] + [itex]\frac{n!}{(n-1)!(n-(n-1))}[/itex]n[itex]^{n-1}[/itex]+[itex]\frac{n!}{(n-2)!(n-(n-2))}[/itex]n[itex]^{n-2}[/itex] + [itex]\cdots[/itex]

= (1 + n)[itex]^{n}[/itex] + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?
Yes, that's correct, and probably what you are looking for.
 
[itex]\frac{n!}{0!(n-0)!}[/itex]1[itex]^{n}[/itex] + [itex]\frac{n!}{(n!)(n-n)!}[/itex]n[itex]^{n}[/itex] + [itex]\frac{n!}{(n-1)!(n-(n-1))}[/itex]n[itex]^{n-1}[/itex]+[itex]\frac{n!}{(n-2)!(n-(n-2))}[/itex]n[itex]^{n-2}[/itex] + [itex]\cdots[/itex]

= 1 + n[itex]^{n}[/itex] + and this is where I get stuck

Am I on the right path by the way in solving the orginal problem?
What would the coefficients of n-1 and n-2 be then?
 
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What would the coefficients of n-1 and n-2 be then?
You mean the coefficients of nn-1?

From your previous post..

[tex]\frac{n!}{(n-1)!(n-(n-1))!}n^{n-1}[/tex]
 
Yes that's correct many thanks
 

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