Question

Let $n$ be a positive integer and define $f\left(n\right)=1!+2!+3!+\cdots +n!$.Find Polynomials $P\left(x\right)$ and $Q\left(x\right)$ such that $f(n+2)=Q\left(n\right)f\left(n\right)+P\left(n\right)f(n+1)$ forall $n\ge 1$
Find P(2).

Answer 1

$f\left(n\right)=1!+2!+3!+\cdots +n!$

$f(n+2)=Q\left(n\right)f\left(n\right)+P\left(n\right)f(n+1)$

$1!+2!+\cdots +(n+2)!=Q\left(n\right)[1!+2!+\cdots +n!]+P\left(n\right)[1!+2!+\cdots +(n+1)!]$

Comparing coefficients upto $(n-1)!$

$Q\left(n\right)+P\left(n\right)=1⟹Q\left(n\right)=1-P\left(n\right)$

$n!+(n+1)!+(n+2)!=(n!)Q\left(n\right)+(n!+(n+1)!)P\left(n\right)$

$1+n+1+n^2+3n+2=1-P\left(n\right)+(n+2)P\left(n\right)$

$P\left(n\right)=n+3⟹Q\left(n\right)=-(n+2)$

$P\left(2\right)=5$