Suppose we have a sequence $\{a_0,a_1,…,a_{N-1}\}$ where $N$ is a large number and we have the following properties:
$$a_0=\frac{1}{N! 4^N}$$
$$a_{q}(4N+2q+1)-2 a_{q-1}=\frac{(-1)^q \left(4N-2q+1\right)}{(N-q)!(2q+1)!}\frac{1}{4^{N-q}} {\ \rm for\ } q=1,..,N-1.$$
Then how to derive the general term formula from the above iterative relation?
How to Derive the General Term Formula for a Sequence
sequences-and-series
Best Answer
Your sequence constitutes a first-order nonhomogeneous linear recurrence relation problem with non-constant coefficients, which could be restated as $$ \left\{ \begin{array}{l} \alpha_q a_q - 2a_{q-1} = \beta_q \\ a_0 = \frac{1}{4^N N!} \end{array} \right. \verb+ +\mathrm{where}\verb+ + \left\{ \begin{array}{l} \alpha_q = 4N+2q+1 \\ \beta_q = \frac{(-1)^q}{4^{N-q}} \frac{4N-2q+1}{(N-q)!(2q+1)!} \end{array} \right. $$
One can get rid of the non-constant coefficient $\alpha_q$ by multiplying the recurrence relation by the factor $\displaystyle\prod_{n=1}^{q-1} \alpha_n$, such that the problem now becomes : $$ b_q - 2b_{q-1} = \gamma_q \verb+ +\mathrm{where}\verb+ + \left\{ \begin{array}{l} b_q = a_q \displaystyle\prod_{n=1}^q \alpha_n \\ \gamma_q = \beta_q \displaystyle\prod_{n=1}^{q-1} \alpha_n \end{array} \right. $$
In the same way as linear differential equations, the general solution of this nonhomogeneous recurrence relation is composed of homogeneous and particular solutions in accordance with the superposition principle, i.e. $b_q = b_q^H + b_q^P$.
The homogeneous part satisfies $b_q^H - 2b_{q-1}^H = 0$, which trivially gives $b_q^H = 2^q b_0^H$.
Again, the particular solution can be constructed thanks to a "variation of parameter" method. Beginning with $b_q^P = 2^q c_q$, one has $b_q^P - 2b_{q-1}^P = 2^q (c_q - c_{q-1}) = \gamma_q$. The new sequence $c_q$ is simply found thanks to the following telescoping series :
$$ c_q - c_0 = \sum_{k=1}^{q} (c_k-c_{k-1}) = \sum_{k=1}^{q} \frac{\gamma_k}{2^k} $$ such that $$ b_q^P = 2^q \left(c_0 + \sum_{k=1}^{q} \frac{\gamma_k}{2^k}\right) $$ hence $$ b_q = b_q^H + b_q^P = 2^q \left(A + \sum_{k=1}^{q} \frac{\gamma_k}{2^k}\right), $$ where $A = b_0^H + c_0$ is just the "constant of integration", and finally one gets $$ a_q = \frac{b_q}{\displaystyle\prod_{n=1}^q \alpha_n} = \frac{2^q}{\displaystyle\prod_{n=1}^q \alpha_n} \left(A + \sum_{k=1}^{q} \frac{\gamma_k}{2^k}\right) $$
The initial condition implies $A = a_0$. The final solution is thus, after substituting back your initial parametrization, $$ a_q = \frac{2^q}{\displaystyle\prod_{n=1}^q (4N+2n+1)} \left(\frac{1}{4^N N!} + \sum_{k=1}^{q} \frac{1}{2^k} \frac{(-1)^k}{4^{N-k}} \frac{4N-2k+1}{(N-k)!(2k+1)!} \prod_{j=1}^{k-1}(4N+2j+1)\right) $$