which one is bigger $100^n+99^n$ or $101^n$

Of course $$1.01^n>1+\frac{n}{100}>2$$ for $n>100$, and obviously $1+0.99^n<2$.

What happened when you tried binomial expansion?

$101^n = (100 + 1)^n = 100^n + n*100^{n-1} + ....$

So you said $n > 1000$ (!!!) so

$100^n + n*100^{n-1}> 100^n + 1000*100^{n-1} > 100^n + 100*100^{n-1} = 100^n + 100^n> 100^n + 99^n$.

By quite a bit! I'm not sure how you could have missed that if you tried the binomial expansion.

The binomial expansion of $ (k+1)^n$ starts:

$$ (k+1)^n = k^n+nk^{n-1}+\cdots$$

so $ (k+1)^n > k^n+nk^{n-1}$ (with $k,n>1$).

With $n=99, $ we have:

$$\begin{align} 101^{99} &> 100^{99} + 99\cdot 100^{98}\\ &> 100^{99} + 99^{99} \\ \end{align}$$

Pushing this technique a little further, and still just using $ (k+1)^n > k^n+nk^{n-1},$ we can take $n=62, $ to get:

$$\begin{align} 101^{62} &> 100^{62} + 62\cdot 100^{61}\\ &> 100^{62} + 62(99^{61} + 61\cdot99^{60})\\ &\quad= 100^{62} + 62\cdot 99^{61} + 3782\cdot 99^{60}\\ &> 100^{62} + 62\cdot 99^{61} + 37\cdot 99\cdot 99^{60}\\ &\quad= 100^{62} + (62+37)\cdot 99^{61}\\ \text{so}\quad101^{62} &> 100^{62} + 99^{62}\\ \end{align} $$

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And of course $101^k>100^k+99^k \implies 101^{k+1} > 101(100^k+99^k) > 100^{k+1}+99^{k+1}$