Given that f(n)= n.f(n-1)+ 2(n+1)! and given that f(0)=1, we can write the solution as: Given that f(n)= n.f(n-1)+ 2(n+1)! and given that f(0)=1, we can write the solution as:
Select one:
a. f(n)= n!(n(n+2) + 1)
b. f(n)=n!(n(n+3)/2 + 1)
c. f(n)= n!(n(n+3)+1)
d. f(n)= n(n+3) + 1
So I am used to finding recurrence equations from questions such as f(n) = f(n-1) + f(n-2). However when I look at this question it puts me off a little because the layout is awkward.
Why do we need f(0) = 1 in the question? it seems like they express f(n) in terms of only n.
I also find it a little odd, because when I put f(0) back into the equation f(n)= n.f(n-1)+ 2(n+1)! I get,  f(0) = 0.f(-1) + 2(1)! = 2 , which clearly is not the case, as f(0) = 1. Am I doing something wrong?
How would do I express f(n) = n.f(n-1) + 2(n+1)! in terms of only n?
 A: Let $f(n)=g(n)\cdot n!$
$n=0\implies\cdots g(0)=1$
$g(n)\cdot n!=n g(n-1)\cdot (n-1)!+2(n+1)!$
$\iff g(n)=g(n-1)+2(n+1)$
Let $g(n)=h(n)+an^2+bn+c\  \ \ \  (1)$
$1=g(0)=h(0)+c\iff h(0)=1-c$
$\implies h(n)+an^2+bn+c=h(n-1)+a(n-1)^2+b(n-1)+c+2(n+1)$
$\iff h(n)=h(n-1)+a-b+2+2n(1-a)$
Set $1-a=0\iff a=1,a-b+2=0\iff b=a+2=3$
to get $h(n)=h(n-1)=\cdots=h(0)=1-c$
From $(1),g(n)=1-c+(1)n^2+(3)n+c=n^2+3n+1$
A: Assuming you mean $2(n+1)!$, if you divide LHS and RHS through $n!$, then set $b(n) = \frac{f(n)}{n!}$, you get 
$$
b(n) = b(n-1) + 2(n+1)
$$
and you life becomes much easier! 
A: Solving such a multiple choice problem does not require solving the recurrence. More simply, it requires only computing enough initial values to exclude the nonsolutions. Here the value of $\,f(1)\,$ excludes all but $2$ choices and the value of $f(2)$ narrows it down to a unique candidate. This can be done in a minute or two of mental arithmetic.
You need the value of $f(0)$ because this is the base case of the recursive definition of $\,f(n).\,$ From that, using the recurrence, you can (uniquely) calculate the value of $\,f(n)\,$ for all positive integers.
