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Homework Help: Binomial theorem to prove

  1. Mar 31, 2012 #1
    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 !!
  2. jcsd
  3. Mar 31, 2012 #2


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    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.
  4. Mar 31, 2012 #3
    Will do that and write my results here. What I don't know is how to get last terms power to n-1 ?
  5. Mar 31, 2012 #4
    I'll be doing this tomorrow so please check ur post.
  6. Apr 1, 2012 #5
    (1+n)^n=1+n/1 n+(n(n-1) n^2)/2!+(n(n-1) 〖(n-2)n〗^3)/3!…
    Last edited: Apr 1, 2012
  7. Apr 1, 2012 #6


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    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.
  8. Apr 2, 2012 #7
    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
  9. Apr 2, 2012 #8


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    Yes. That is why there is a ##\ge## sign in the statement of the problem. Leaving out the other terms makes it smaller.
  10. Apr 2, 2012 #9
    thank you- this forum is blessing to someone relatively new to pure maths.
  11. May 21, 2012 #10
    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
  12. Jun 12, 2012 #11
    what do you mean last 3 terms? if its infinite it wouldn't have any last term?
  13. Jun 12, 2012 #12


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    ##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##.
  14. Jun 13, 2012 #13
    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.
  15. Jun 13, 2012 #14
    They're from the same formula. Set a=1, b=x.
  16. Jun 13, 2012 #15
    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?

  17. Jun 13, 2012 #16
    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??
  18. Jun 13, 2012 #17


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    \binom n {n-1}=\frac {n!}{(n-1)!(n-(n-1))!}=\frac {n!}{(n-1)!(n-n+1)!}
  19. Jun 14, 2012 #18
    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]
  20. Jun 14, 2012 #19
    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??
  21. Jun 14, 2012 #20
    [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?
  22. Jun 14, 2012 #21
    Yes, that's correct, and probably what you are looking for.
  23. Jun 19, 2012 #22
    What would the coefficients of n-1 and n-2 be then?
  24. Jun 19, 2012 #23
    You mean the coefficients of nn-1?

    From your previous post..

  25. Jun 20, 2012 #24
    Yes that's correct many thanks
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