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Consider the following recurrence relation for $C_i(r)$s

$$\begin{align} &C_0(r)=r-r_2 \\ &q(r+n)C_n(r)+\sum_{k=0}^{n-1}[(r+k)\alpha_{n-k}+\beta_{n-k}]C_k(r)=0, \qquad n\ge1 \end{align} \tag{1}$$

where $r$ is some real variable, $\alpha_i$ and $\beta_i$ are some arbitrary real sequences and $q(r)$ is given by

$$q(r)=(r-r_1)(r-r_2) \tag{2}$$

Also, $r_1 \gt r_2$ are two integer roots of $q(r)=0$ such that $r_1-r_2=N.$

I want to prove that

$$C_{n+N}(r_2)=C_n(r_1)C_N(r_2), \qquad n=1,2,3,... \tag{3}$$

This problem arises in the Frobenius method for solving second order linear ordinary differential equations with variable coefficients when one tries to show the linear dependency of two solutions $y_1=x^{r_1}\sum_{n=0}^{\infty}C_n(r_1) x^n$ and $y_2=x^{r_2}\sum_{n=0}^{\infty}C_n(r_2) x^n$.

Any hint or help is appreciated. :)

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We can see that for $r=r_1$ by Eq.$(1)$ we have

$$\begin{align} &C_0(r_1)=r_1-r_2=N \\ &q(r_1+n)C_n(r_1)+\sum_{k=0}^{n-1}[(r_1+k)\alpha_{n-k}+\beta_{n-k}]C_k(r_1)=0, \qquad n\ge1 \end{align} \tag{4}$$

and hence by back substitution, using Eq.$(4)$, we can find $C_n(r_1)$s. Focusing on $r=r_2$, we will see that

$$\begin{align} &C_0(r_2)=r_2-r_2=0\\ &q(r_2+n)C_n(r_2)+\sum_{k=0}^{n-1}[(r_2+k)\alpha_{n-k}+\beta_{n-k}]C_k(r_2)=0, \qquad n\ge1 \end{align} \tag{5}$$

and hence it can be concluded that

$$C_n(r_2)=0, \qquad n=0,1,2,...,N-1 \tag{6}$$

but when $n=N$, noticing that $q(r_2+N)=q(r_1)=0$ then we encounter

$$0 \times C_N(r_2) = 0 \tag{7}$$

and hence $C_N(r_2)$ can be chosen to be a non-zero constant such as $C$. So, re-writing Eq.$(5)$ according to Eq.$(6)$, we obtain

$$\begin{align} &C_N(r_2) = C\\ &q(r_2+n)C_{n}(r_2)+\sum_{k=N}^{n-1}[(r_2+k)\alpha_{n-k}+\beta_{n-k}]C_k(r_2)=0, \qquad n \ge N+1 \end{align} \tag{8}$$

then replacing $n$ by $n+N$ in Eq.$(8)$ we get

$$\begin{align} &C_N(r_2) = C\\ &q(r_2+N+n)C_{n+N}(r_2)+\sum_{k=N}^{n+N-1}[(r_2+k)\alpha_{n+N-k}+\beta_{n+N-k}]C_k(r_2)=0, \qquad n \ge 1 \end{align} \tag{9}$$

next, noting that $r_1=r_2+N$ and replacing $k$ by $k+N$ in the summation in Eq.$(9)$ we get

$$\begin{align} &C_N(r_2) = C\\ &q(r_1+n)C_{n+N}(r_2)+\sum_{k=0}^{n-1}[(r_1+k)\alpha_{n-k}+\beta_{n-k}]C_{k+N}(r_2)=0, \qquad n \ge 1 \end{align} \tag{10}$$

but $(10)$ is just the same recurrence relation as $(4)$ with different initial term. So we can conclude that

$$\frac{C_{n+N}(r_2)}{C_N(r_2)}=\frac{C_n(r_1)}{C_0(r_1)} \tag{11}$$

if we choose $C_0(r_1)=1$ then Eq.$(11)$ turns into Eq.$(3)$.

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