$t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, find the closest pair of integers, a and b, such that, $a \lt t \lt b$. $t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$
find the closest pair of integers, a and b, such that, $a \lt t \lt b$.
$30=1+29$
$(1+29)^{65}=(1+29)(1+29)^{64}$
 A: I actually found this easier to do without algebra. The numerator looks like $30*$ denominator + error term, so you can write 
$$\frac{30^{65}-29^{65}}{30^{64}-29^{64}} = \frac{30^{65} - 30*29^{64} + 30*29^{64} - 29^{65}}{30^{64} - 29^{64}}$$
$$= 30 + \frac{30*29^{64} - 29^{65}}{30^{64} - 29^{64}}$$
$$= 30 + \frac{30*29^{64} - 29*29^{64}} {30^{64} - 29^{64}}$$
$$= 30 + \frac{29^{64}}{30^{64} - 29^{64}}$$
The idea is that the denominator here should be much larger than the numerator, so the overall fraction is between $30$ and $31$. There are a few ways to show this. One is to expand $30^{64} = (29 + 1)^{64}$ via the binomial theorem, obtaining
$$30^{64} = 29^{64} + 64*29^{63} + ...$$
Thus the denominator $30^{64} - 29^{64}$ is greater than $64*29^{63}$, which in turn is greater than the numerator $29^{64}$ since $64 > 29$. Thus indeed the second term is less than $1$, so that the overall fraction is between $30$ and $31$.
A: $t=30+(29^{64}/d)$ with $d=30^{64}-29^{64}$. Since $30^{64}=29^{64}(1+1/29)^{64}\gt29^{64}\,(1+64/29)$, $d\gt29^{64}\cdot(64/29)$ and $30\lt t\lt30+(29/64)\lt31$.
Likewise, $s=((a+1)^{n+1}-a^{n+1})/((a+1)^n-a^n)$ yields $a+1\lt s\lt a+1+(a/n)$ hence $a+1\lt s\lt a+2$ as long as $a\lt n$.
A: Let $A,B,a,b,n$ be positive numbers.
$\frac{A^{n+1}-B^{n+1}}{A^n-B^n}-A=\frac{B^n(A-B)}{A^n-B^n}$ will be $>0$ if $A>B$ and $n ≥ 1$
Putting $A=a+1,B=a$, $\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}>(a+1)$ for $n ≥ 1$.
Alternatively, 
$\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}-(a+1)$
$=\frac{(a+1)^{n+1}-a^{n+1}-(a+1)^{n+1}+(a+1)a^n}{(a+1)^n-a^n}$
$=\frac{a^n}{(a+1)^n-a^n}>0$ for $n ≥ 1$
$=>\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}>(a+1)$  for $n ≥ 1$
Now, $\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}-(a+1+1)$
$=\frac{a^n}{(a+1)^n-a^n}-1 $
$=\frac{2a^n-(a+1)^n}{(a+1)^n-a^n}$
$=\frac{2-(1+\frac{1}{a})^n}{(1+\frac{1}{a})^n-1}$
But $(1+\frac{1}{a})^n=1+n\cdot\frac{1}{a}+\frac{n(n-1)}{2}\frac{1}{a^2}+...$ will be $>2$  if $n>a$
So, $2-(1+\frac{1}{a})^n<0$   if $n>a$
$\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}<(a+1+1)$ if $n>a$
So, $(a+1)<\frac{(a+1)^{n+1}-a^{n+1}}{(a+1)^n-a^n}<(a+1+1)$ if $n>a$
Here in this problem, $n=64,a=29$
