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

And of course $101^k>100^k+99^k \implies 101^{k+1} > 101(100^k+99^k) > 100^{k+1}+99^{k+1}$
A: $100^n + 99^n$ and  $101^n$
Divide all terms by $101^n$
$\left(\dfrac{100}{101}\right)^n+\left(\dfrac{99}{101}\right)^n<2 \left(\dfrac{100}{101}\right)^n$
and $2 \left(\dfrac{100}{101}\right)^n<1$ when $ \left(\dfrac{100}{101}\right)^n<\dfrac{1}{2}$
that is when $n\log\dfrac{100}{101}<\log\dfrac{1}{2}$
$n\geq 70$
so for $n>1000$  we have $100^n + 99^n < 101^n$
A: Of course
$$1.01^n>1+\frac{n}{100}>2$$
for $n>100$, and obviously $1+0.99^n<2$.
A: 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 $0\le n\le 48$) 
A: The left hand side of the following beomes monotonic decreasing after a certain n>m.
$(\frac{100}{101}) ^{n} +  (\frac{99}{101} )^{n} <1 \;\; \forall  n>m$
