Results 1 to 4 of 4

Math Help - A solution without Newton's binom?

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    2

    A solution without Newton's binom?

    Demonstrate:

    (2+a)^k+(2-a)^k <=4^k

    Thank You.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    More information please

    Hello ira069

    Welcome to Math Help Forum!
    Quote Originally Posted by ira069 View Post
    Demonstrate:

    (2+a)^k+(2-a)^k <=4^k

    Thank You.
    I think we need a bit more information here.

    This statement clearly isn't true for all values of a and k. For instance, if a = 3 and k is an integer, (2+a)^k+(2-a)^k = 5^k +(-1)^k = 5^k \pm 1 depending upon whether k is even or odd. Clearly for k>1, this is greater than 4^k.

    So I'm not really sure what the question means.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2009
    Posts
    2

    Thank you.

    Thank you.

    I am asking for a solution without Newton's binom,
    for k integer positive, -2 < a < 2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008
    Quote Originally Posted by ira069 View Post
    Thank you.

    I am asking for a solution without Newton's binom,
    for k integer positive, -2 < a < 2.
    Since k is a positive integer, you don't need Newton's Binomial Theorem, you just need the integer Binomial Theorem.


    Remember that (x + y)^n = \sum_{r = 0}^{n}{\left(_r^n\right)x^{n - r}y^r}

    So (2 + a)^k = \sum_{r = 0}^k{\left(_r^k\right)2^{k - r}a^r} and (2 - a)^k = \sum_{r = 0}^k{\left(_r^k\right)2^{k - r}(-a)^r}

    Therefore (2 + a)^k + (2 - a)^k = \sum_{r = 0}^k{\left(_r^k\right)2^{k - r}a^r} + \sum_{r = 0}^k{\left(_r^k\right)2^{k - r}(-a)^r}

    Try writing these sums in expanded form and you should be able to prove your original statement.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 23rd 2011, 03:39 AM
  2. \binom in a table?
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: February 19th 2011, 03:42 AM
  3. Efficiency of p in binom(n,p)
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: September 15th 2010, 06:18 PM
  4. Replies: 1
    Last Post: March 24th 2010, 12:14 AM
  5. Replies: 2
    Last Post: September 7th 2009, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum