-1

Prove for $\ n\ge100,n\ge50,n\ge49 \ $that the following is true:

$$\ 101^{n}>100^{n}+99^{n} \ $$

Prove for $\ n\ge100,n\ge50,n\ge49 \ $

I know it’s somehow connected with Bernoulli; however, I'm having trouble figuring out how to do this.

Infinity00
  • 127
Valerie
  • 33
  • The inequality is true for $n=49$, but it's false for $n=50$ and $n=100$. In fact, $101^{50}-100^{50}-99^{50} = -3022603911549554928$ and $101^{100}-100^{100}-99^{100} = -831363304571293664$. – GNUSupporter 8964民主女神 地下教會 Mar 01 '18 at 22:54
  • Yeah, but how can you prove it without calculation? – Valerie Mar 01 '18 at 22:58
  • @GNUSupporter Are you sure? I think $101^{50}-100^{50}-99^{50}$ may be about $+3.96\times 10^{98}$ and it only gets bigger from there. – Henry Mar 01 '18 at 22:59
  • My point is that you can't prove something ($n = 50, 100$) that is false. – GNUSupporter 8964民主女神 地下教會 Mar 01 '18 at 22:59
  • @GNUSupporter If it is true for $n$ it is true for anything greater than $n$. To see that, write it as $1≥\left( \frac {100}{101}\right)^n+\left( \frac {99}{101}\right)^n$ – lulu Mar 01 '18 at 23:00
  • It is totally right for all the conditions, but I need to prove it using some theory. – Valerie Mar 01 '18 at 23:00
  • 1
    I see it as false for $n=48$ but true for all $n≥49$. – lulu Mar 01 '18 at 23:00
  • @Henry ,lulu OIC. Sorry for my mistake. – GNUSupporter 8964民主女神 地下教會 Mar 01 '18 at 23:01
  • I don't see how to avoid the calculation for $n=49$. Writing it my way, with $n=48$ the right hand is $1.003141$ and at $n=49$ the right hand is $0.989418$. It's very close! – lulu Mar 01 '18 at 23:02
  • @lulu can’t we use induction? – Valerie Mar 01 '18 at 23:03
  • Induction has to start somewhere. Like I say, once you know it for $n=49$ you are done. But I don't see a way to make it obvious for $49$ ...which doesn't mean there isn't one, of course. I mean, writing it as you did, we see that $101^{49}-100^{49}-99^{49}\approx 1.72\times 10^{96}$ which certainly makes it look less close than my method does. – lulu Mar 01 '18 at 23:06
  • @lulu I also tried to use the binomial theorem and Bernoulli inequality, but it didn’t work out. – Valerie Mar 01 '18 at 23:08
  • There could be a proof if you could show $e^{49/101} \gt \frac{1+\sqrt{5}}{2}$, which is in fact true – Henry Mar 01 '18 at 23:11
  • The binomial theorem might work. Write $101^{49}=(100+1)^{49},99^{49}=(100-1)^{49}$. – lulu Mar 01 '18 at 23:11
  • @Henry how exactly can I show it? – Valerie Mar 01 '18 at 23:12
  • I can get the binomial theorem to do it for $n=50$...not instantly clear that I can tighten the method to get $n=49$. I'll write it up below. – lulu Mar 01 '18 at 23:14
  • @Valerie: no idea without a calculator, but it would imply $e^{-49/101} +e^{-2 \times 49/101} \lt 1 $ and thus $\left(1- \frac {1}{101}\right)^{49}+\left(1 -\frac {2}{101}\right)^{49} \lt 1$ – Henry Mar 01 '18 at 23:16
  • Just added an argument to get $n=49$. It's significantly messier and should definitely be checked for arithmetic blunder. – lulu Mar 01 '18 at 23:33
  • @lulu yeah, thank you very much. Great that some of my thoughts are actually not that pointless. I’ll see what my professor thinks about it tomorrow. – Valerie Mar 01 '18 at 23:38
  • Good luck. $\quad $ – lulu Mar 01 '18 at 23:40

1 Answers1

3

I'll show it for $n=50$. That will, of course, show it for $n≥50$ though not immediately for $n=49$.

Writing $101=100+1,99=100-1$ we see that $$101^{50}-99^{50}=\sum_{i=0}^{50}\left(1-(-1)^i\right)\binom {50}i100^{50-i}=2\sum_{k=0}^{24}\binom {50}{2k+1}100^{50-2k-1} $$

Now, the first term in the sum is $$\binom {50}1100^{50}=50\times 100^{49}=\frac 12\times 100^{50}$$ so we are done.

Perhaps some tightening of this will suffice for $n=49$, but I did not try. As stated in the comments, the claim is false if $n=48$.

Added: The same method works for $n=49$ but you need the first two terms in the sum. Specifically, the same method shows that $$101^{49}-99^{49}=\sum_{i=0}^{49}\left(1-(-1)^i\right)\binom {49}i100^{49-i}=2\sum_{k=0}^{24}\binom {49}{2k+1}100^{49-2k-1} $$ Now, the first term in the sum is $$49\times 100^{48}$$ and the second term is $$\binom {49}3\times 100^{46}=18424\times 100^{46}>100^2\times 100^{46}=100^{48}$$ and adding these two terms completes the argument as before.

lulu
  • 70,402