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Math Help - A solution without Newton's binom?

  1. #1
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    A solution without Newton's binom?

    Demonstrate:

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

    Thank You.
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  2. #2
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    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
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  3. #3
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    Thank you.

    Thank you.

    I am asking for a solution without Newton's binom,
    for k integer positive, -2 < a < 2.
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  4. #4
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    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.
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