In the book concrete maths (2nd ed, page 27). The author explains of a way to solve recurrence in the following manner
Consider a general recurrence of form
$a_n T_n = b_n T_{n-1} + c_n$
Multiply $s_n$ to both sides
$s_n a_n T_n =s_n b_n T_{n-1} + s_n c_n$
such that $s_{n-1} a_{n-1} = s_n b_n$ — eq(1)
substituting values
$s_n a_n T_n =s_{n-1} a_{n-1} T_{n-1} + s_n c_n$
Let $s_n a_n T_n = F_n$
$F_n = F_{n-1} + s_n c_n$
Hence $$F_n = s_0 a_0 T_0 + \sum_{k=1}^{n} s_k c_k$$
To choose $s_n$, consider equation 1
$$s_{n-1} a_{n-1} = s_{n} b_n$$
$$s_n = \frac{s_{n-1} a_{n-1}}{ b_n}$$
Expanding $s_{n-1}$
$$s_n = \frac{a_{n-1} a_{n-2} a_{n-3} ….. a_{1}}{b_n b_{n-1} b_{n-2} … b_{2}}$$
Now this is the part that gets confusing, shouldn't there be a $s_1$ in conjugation with $a_1$.
For example, consider $n=3$
$$s_3 = \frac{s_2 a_2}{b_3}$$
$$s_3 = \frac{\frac{s_1 a_1}{b_2} a_2}{b_3}$$
$$s_3 = \frac{s_1 a_1 a_2}{b_2 b_3}$$
Image of relevant text in the book
Best Answer
You're absolutely right. Sometimes in equations like this the first term $s_1$ disappears simply because it is equal to $1$, but it doesn't seem to be the case here.