Hi, I can't seem to understand this. I do understand that (a^n - b^n) is divisible by (a - b), and n belongs to the set of natural numbers. But the way it has been proved in the book, it just doesn't make any sense to me. How does the series (it's a G.P) prove that statement? Please explain it to me. Thanks
On
For some reason the book sets $a$ apart.
I suspect that was done because on forehand it was proved already that $$\frac{1-x^n}{1-x}=1+x+\dots+x^{n-1}$$which is used in the proof.
Personally I don't like it.
This because from:
$$(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}=\sum_{k=0}^{n-1}a^{n-k}b^{k}-\sum_{k=0}^{n-1}a^{n-k-1}b^{k+1}=\sum_{k=0}^{n-1}a^{n-k}b^{k}-\sum_{k=1}^{n}a^{n-k}b^{k}=a^n-b^n$$it follows immediately that $a-b$ divides $a^n-b^n$. Further it avoids fractions and the setting apart of $a$. The only thing against it is that there is no searching for a quotient, but makes use of a conjecture.
Next it follows immediately that $a+b=a-(-b)$ divides $a^n-(-b)^n=a^n+b^n$ if $n$ is odd.
On
Let $r:=\dfrac ba$.
A first step is to eliminate $b$:
$$\frac{a^n-b^n}{a-b}=\frac{a^n}a\frac{1-\dfrac{b^n}{a^n}}{1-\dfrac ba}=a^{n-1}\frac{1-r^n}{1-r}.$$
Then you should recognize the summation of a geometric series and write
$$=a^{n-1}(1+r+r^2+\cdots+r^{n-1})$$ which is also
$$=a^{n-1}\left(1+\frac ba+\frac{b^2}{a^2}+\cdots+\frac{b^{n-1}}{a^{n-1}}\right).$$
And finally
$$=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots b^{n-1}.$$
On
It's easy to believe you don't understand this proof.
If you have already proved that $$ 1+x+\dots+x^{n-1}=\frac{1-x^n}{1-x} \tag{*} $$ which seems necessary for the present proof, just set $$ x=\frac{b}{a} $$ Then, after substituting and collecting denominators in the left-hand side of ($\text{*}$), we have $$ 1+\frac{b}{a}+\frac{b^2}{a^2}+\dots+\frac{b^{n-1}}{a^{n-1}} = \frac{a^{n-1}+ba^{n-2}+\dots+b^{n-1}}{a^{n-1}} $$ On the other hand, substituting in the right-hand side of ($\text{*}$) yields $$ \frac{1-\dfrac{b^{n}}{a^{n}}}{1-\dfrac{b}{a}} =\frac{\dfrac{a^n-b^n}{a^n}}{\dfrac{a-b}{a}}= \frac{a^n-b^n}{a^n}\frac{a}{a-b}= \frac{1}{a^{n-1}}\frac{a^n-b^n}{a-b} $$ Hence $$ \frac{a^{n-1}+ba^{n-2}+\dots+b^{n-1}}{a^{n-1}} = \frac{1}{a^{n-1}}\frac{a^n-b^n}{a-b} $$ and removing the common term $a^{n-1}$ finally gives $$ a^{n-1}+ba^{n-2}+\dots+b^{n-1}=\frac{a^n-b^n}{a-b} $$
You can (in second step) see the fraction converting to the series if you try polynomial division of $\frac{1-t^n}{1-t}$ and try and find its quotient.