$t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, 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, $a \lt t \lt b$.
$30=1+29$
$(1+29)^{65}=(1+29)(1+29)^{64}$

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, $n=64,a=29$
