Simplify recurrence relations with known solution form

Question

If I have a linear recurrence relation

$$Q_{n+1}(X) = \int_0^X Q_n (x) e^{k (X-x)} \ \mathrm{d} x,$$

knowing $Q_0(x)=e^{kx}$ and $k$ is a constant. It is easy to show the general form of the solution is

$$Q_n(x) = e^{kx} \sum_{i=0}^n \alpha_{i,n} x^i,$$

and we can establish a recurrence relation for the coefficients $\alpha_{i,n}$.

The aim is to write a code in Wolfram Language that can find the recurrence relation for the coefficients $\alpha_{i, n}$ in this scenario, and in slightly more general scenarios (linear recurrence relation involving a single integral and similar solution form).

With ReplaceAll and properly defined Rules, I am able to insert the general solution form into the original recurrence relation, but I am unable to simplify further. For instance, Mathematica doesn't pull the sum out of the integral in the RHS.

Update 1

The coefficients $\alpha_{i,n}$ should also have an $n$ index.

Update 2

Indeed we know how to solve this recurrence equation explicitly but that is not the goal here. For human beings, it is rather simple to insert the general form of the solution, then exchange the sum and the integral in the RHS, also pulling out the coefficients $\alpha_{i,n}$ of the integral as they have no $x$ dependence. What I want to achieve is to write a code in Wolfram Language that does it in the way human beings would do.

Answer 1

Regarding the integral equation

$$ Q_{n+1}(X) = \int_0^X Q_n (x) e^{k (X-x)} \ \mathrm{d} x, $$

it can be written as

$$ Q_{n+1}(x) = Q_n(x)\circledast e^{kdx} $$

and taking the Laplace transform we have

$$ Q_{n+1}(s)=\frac{1}{s+k}Q_{n}(s) $$

now considering that $Q_0(s) = \frac{1}{s+k}$ we have

$$ Q_n(s) = \left(\frac{1}{s+k}\right)^{n+1} $$

with anti-transform

$$ Q_n(x) = \frac{e^{-k x} x^n}{\Gamma (n+1)} $$

Follow two scripts which produce the same results

Clear[Q]
InverseLaplaceTransform[Q[n] /. RSolve[{Q[n + 1] == Q[n]/(s + k), Q[0] == 1/(s + k)}, Q, n][[1]], s, x]

and

Clear[Q]
n = 10;
Q = Exp[k x];
QQ = {Q};
For[i = 1, i <= n, i++,
 Q = Integrate[Q Exp[k (u - x)], {x, 0, u}] /. {u -> x};
 AppendTo[QQ, Q]
 ]

QQ

Answer 2

Try NestList

sol = NestList[
Function[X,Evaluate@Integrate[ # [x] Exp[k (X - x)  ], {x, 0, X}]] &    , 
Exp[k #] &, 5];

Qn=Through[sol[X]]
(*{E^(k X), E^(k X) X, 1/2 E^(k X) X^2, 1/6 E^(k X) X^3,1/24 E^(k X) X^4,1/120 E^(k X) X^5}*)    

It looks like your expectation might be wrong. Mathematica evaluates instead

Q[n][X]=Exp[k X] X^n/n!

This result might be confirmed with (Thanks @Mariusz Iwaniuk)

FindSequenceFunction[Qn, n] // FunctionExpand