# $t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, 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, $a \lt t \lt b$.

$30=1+29$

$(1+29)^{65}=(1+29)(1+29)^{64}$

-
Hint: divide both the denominator and numerator by $29^{64}$. – Frenzy Li Aug 21 '12 at 7:00

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 $31$.

-
This is exactly what i was going to post +1. – Aang Aug 21 '12 at 14:32

$t=30+(29^{64}/d)$ 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 $a\lt n$.

-

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, $n=64,a=29$

-