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It is given that $a_{n+1}=(n+1)a_n+2(n+1)!$ where $n>0$ and and $a_0=0$ , so find the explicit formula for $a_n$.

I tried to use exponential generating functions such that , lets say $A(x)= \sum_{n=0}^{\infty}a_n\frac{x^n}{n!}=(e^x-1)$ ,because $a_0=0$ .Then:

  • $$a_{n+1}=\frac{n(A(x)-a_0)}{x}=\frac{nA(x)}{x}$$

  • $$(n+1)a_n=nA(x)+A(x)$$

  • $$2(n+1)!=2(n+1)\sum_{n=0}^{\infty}x^n=\frac{2n+2}{1-x}$$

As a result , i found that $$\frac{nA(x)}{x}=nA(x)+A(x)+\frac{2n+2}{1-x}$$

However , i could not take the $A(x)$ from the foregoing expression(i.e it does not provide an answer).I guess i am making a silly error but i stuck. So , can you help me to solve this recurrence relation using generating functions ?You do not have to follow my way , other methodds are also acceptable. Thanks in advance !!

Not a Salmon Fish
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    Try to use the recursion you are given to find a recursion for the sequence $b_n = a_n/n!$. It might be a lot easier. – Testcase Mar 21 '22 at 19:22
  • Try working out the first few terms and dividing the results by $n!$. It is easy to spot the pattern and prove it – Henry Mar 21 '22 at 19:24

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Another way to find the formula is by solving

$$ a_n = a_n^h+a_n^p $$

where $a_n^h$ is the solution for

$$ a_n^h = n a_{n-1}^h $$

and $a_n^p$ is a convenient particular solution obeying

$$ a_n^h = n a_{n-1}^h+2n! $$

For the homogeneous recurrence the solution is easily found as

$$ a_n^h = n! c_0 $$

The particular is solved proposing $a_n^p = c_0(n)n!$ and after substitution we get the recurrence

$$ c_0(n) = c_0(n-1) + 2 $$

with particular solution

$$ c_0(n) = 2n $$

and finally we have

$$ a_n = n!c_0 + (2n)n! = (c_0+2n)n! $$

but here $c_0 = a_0 = 0$ hence

$$ a_n = (2n)n! $$

Cesareo
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