Hi. I am trying to prove the following:

1 + C(2n, 1) + C(2n, 2) + ... + C(2n, k) <= (2n)^k for n >= k > 1.

I would appreciate any suggestions you all may have.

Thanks.

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- Aug 26th 2010, 12:07 PMkkoutsothodoroscombinatorial inequality
Hi. I am trying to prove the following:

1 + C(2n, 1) + C(2n, 2) + ... + C(2n, k) <= (2n)^k for n >= k > 1.

I would appreciate any suggestions you all may have.

Thanks. - Aug 26th 2010, 01:15 PMArchie Meade
You could prove it via Proof By Induction...

**P(k)**

$\displaystyle \displaystyle\ 1+\binom{2n}{1}+\binom{2n}{2}+....+\binom{2n}{k}\ \le\ (2n)^k\ ?$

**P(k+1)**

$\displaystyle \displaystyle\ 1+\binom{2n}{1}+\binom{2n}{2}+....+\binom{2n}{k}+\ binom{2n}{k+1}\ \le\ (2n)^{k+1}\ ?$

Try to prove that P(k+1) will be valid if P(k) is valid.

**Proof**

$\displaystyle \displaystyle\ (2n)^k+\binom{2n}{k+1}\ \le\ 2n(2n)^k\ ?$

$\displaystyle \displaystyle\binom{2n}{k+1}\ \le\ (2n-1)(2n)^k$

$\displaystyle \displaystyle\frac{[2n]!}{[2n-(k+1)]![k+1]!}\ \le\ (2n-1)(2n)(2n)(2n)....$

$\displaystyle \displaystyle\frac{(2n)(2n-1)(2n-2)....}{(k+1)!}\ \le\ (2n-1)(2n)(2n)(2n)....$

There are k+1 terms in the numerator of the LHS and k+1 terms on the RHS, since there are k occurances of 2n.

It's easy to see that the inequality holds. - Aug 26th 2010, 01:39 PMkkoutsothodoros
thanks!

- Aug 26th 2010, 02:03 PMArchie Meade
The proof is then complete when you examine the base case for k=2, assuming k is an integer.

$\displaystyle \displaystyle\ 1+\binom{2n}{1}+\binom{2n}{2}\ \le\ (2n)^2\ ?$

$\displaystyle \displaystyle\ 1+2n+\frac{2n(2n-1)}{2}\ \le\ 4n^2\ ?$

$\displaystyle \displaystyle\ 1+2n+n(2n-1)\ \le\ 4n^2\ ?$

$\displaystyle 1+2n+2n^2-n\ \le\ 4n^2\ ?$

$\displaystyle 2n^2+n+1\ \le\ 4n^2\ ?$

$\displaystyle n+1\ \le\ 2n^2\ ?$

$\displaystyle n(2n-1)\ \ge\ 1\ ?$

$\displaystyle \text{\footnotesize\ True for \;\;}n\ \ge\ 1$