Demonstrate:

$\displaystyle (2+a)^k+(2-a)^k <=4^k$

Thank You.

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- Jun 28th 2009, 05:11 AMira069A solution without Newton's binom?
Demonstrate:

$\displaystyle (2+a)^k+(2-a)^k <=4^k$

Thank You. - Jun 28th 2009, 09:54 PMGrandadMore information please
Hello ira069

Welcome to Math Help Forum!I think we need a bit more information here.

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

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

Grandad - Jun 28th 2009, 10:46 PMira069Thank you.
Thank you.

I am asking for a solution without Newton's binom,

for k integer positive, -2 < a < 2. - Jun 28th 2009, 11:48 PMProve It
Since k is a positive integer, you don't need Newton's Binomial Theorem, you just need the integer Binomial Theorem.

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

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

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

Try writing these sums in expanded form and you should be able to prove your original statement.