# A solution without Newton's binom?

• Jun 28th 2009, 06:11 AM
ira069
A solution without Newton's binom?
Demonstrate:

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

Thank You.
• Jun 28th 2009, 10:54 PM
Hello ira069

Welcome to Math Help Forum!
Quote:

Originally Posted by ira069
Demonstrate:

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

Thank You.

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.

• Jun 28th 2009, 11:46 PM
ira069
Thank you.
Thank you.

I am asking for a solution without Newton's binom,
for k integer positive, -2 < a < 2.
• Jun 29th 2009, 12:48 AM
Prove It
Quote:

Originally Posted by ira069
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.