1
$\begingroup$

Here's my question: Let $f(n)=(2 n-1) \times(2 n-3) \times(2 n-5) \times \cdots \times 5 \times 3 \times 1$, and I need to prove that for all $n\ge3,f(n)\ge2\times7^{n-2}$.

I have figured out some interesting points, but still I have no clue how to make use of them.

  1. $f(n+1)=2n\times(2n-2)\times\dots\times4\times2=2^n\times n!$
  2. $f(n+1)\times f(n)=(2n)(2n-1)\dots(3)(2)(1)=(2n)!$

Then everything stops here. I'm trying to make sense with the $7^{n-2}$ with the statement, but failed. So is there any other hints for this question? Thanks a lot for your help!

$\endgroup$
2
  • 2
    $\begingroup$ Use the fact that if $n>3$, then $2n-1>7$. $\endgroup$
    – Peter
    Feb 27 at 10:31
  • $\begingroup$ Note that both your observations are wrong, since $f(n+1) = (2n+1)\times f(n)$, and that is also odd. $\endgroup$
    – Martund
    Feb 27 at 10:58
2
$\begingroup$

Just bound the terms this way: $$f(n)=\underbrace{(2n-1)}_{\geq 7}\underbrace{(2n-3)}_{\geq 7}\ldots\underbrace{(7)}_{\geq 7}(5)\underbrace{(3)}_{\geq 2}$$ and take into account that there are $n$ terms.

$\endgroup$
0
$\begingroup$

we put $\sigma_{2n-1}=(2n-1)!!$ for $n>3$ : We have $\sigma_5=9×7×5×3×2×1>2(7^{5-2})$
and $\sigma_7=9×\sigma_5>7\sigma_5$
By recurence we obtain $\sigma_{2n-1}>7^{n-6}\sigma_5$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.