which one is bigger $100^n+99^n$ or $101^n$ Suppose $n \in \mathbb{N} , n>1000$ now how can we prove :which one is bigger $$100^n+99^n \text{ or  }  101^n \text{ ? }$$ 
I tried to use $\log$ but get nothing . Then I tried for binomial expansion...but  I get stuck on this . 
can someone help me ? thanks in advance.

Consider $$f(x)=100^x+99^x-101^x=101^x\left(\left(\frac{100}{101}\right)^x+\left(\frac{99}{101}\right)^x-1\right)$$ The last factor tends to $-1$ when $x\to\infty$ so $f(x)$ is negative from a certain value of $x$ and since $f(0)=1\gt 0$ and the functions are concave we have 
$$100^x+99^x\gt 101^x\text{ for } x\lt x_0\\100^x+99^x\lt 101^x\text{ for } x\gt x_0$$ where $x_0$ is such that $f(x_0)=0$.
Calculation gives $$48\lt x_0\lt 49$$
Thus $101^n$ is the greater for $n\gt 1000$ 
(Actually for $n\gt48$. Besides  $100^n+99^n$ is greater for $0\le n\le 48$) 
