Prove Bernoulli Task

Question

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.

Answer 1

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.