I'm reading the book Concrete Mathematics and on page 27(chapter sums and recurrences) there is a text I simply cannot understand:
This trick is a special case of a general technique that can reduce virtually any recurrence of the form
$a_nT_n = b_nT_{n-1} + c_n$
To a sum. The idea is to multiply both sides by a summation factor, $S_n$:
$S_na_nT_n = S_nb_nT_{n-1} + S_nc_n$
This factor $S_n$ is cleverly chosen to make
$S_nb_n = S_{n-1}a_{n-1}$ //THIS IS MY FIRST QUESTION, How did they get this result from the previous equation?
Then if we write $S_n=S_na_nT_n$ we have a sum recurrence
$S_n = S_{n-1} + S_nC_n$ //HOW DID THEY GET THIS?
First, you mixed $S_n$ and $s_n$. If you don't make that distincion, the book's explanation won't make any sense. The summation factor you wrote about is $s_n$, lowercase, not $S_n$. You asked about how he concluded that $s_nb_n=s_{n-1}a_{n-1}$. The answer is what MPW said. And its actually written on the book. And you even wrote it here!
"This factor $s_n$ is cleverly chosen to make $s_nb_n=s_{n-1}a_{n-1}$".
If you want a more direct explanation on why he wants $s_n$ to have this property, remember he wants to transform the recurrence into a sum. A recurrence like $F_n=F_{n-1}+c_n$ can be easily transformed into a sum: $F_n= F_0+\sum_{k=1}^{n}c_k$ (calculate $F_n$ for small $n$'s if its not clear for you). He is going to rewrite the original recurrence to look like this. The summation factor $s_n$, which he chose so that it had that property, is multiplied by your original equation:
$s_n(a_nT_n)=s_n(b_nT_{n−1}+c_n) \rightarrow s_na_nT_n=s_nb_nT_{n−1}+s_nc_n $
Since $s_nb_n=s_{n-1}a_{n-1}$ , we have $s_nb_nT_{n−1} = s_{n-1}a_{n-1}T_{n−1}$. So:
$s_na_nT_n=s_{n-1}a_{n-1}T_{n−1}+s_nc_n$
And now he defines a new variable $S_n$ (uppercase) as $S_n=s_na_nT_n$. The recurrence relation becomes $S_n=S_{n-1}+s_nc_n$. With that, we end with a recurrence relation thats easily transformed into a sum, like the $F_n$ I defined before.