Polynomial : $ P(x+1)-2P(x)+P(x-1)=6x $
Let $\delta$ be the operator mapping a polynomial $p(x)$ into the polynomial $(\delta p)(x)=p(x+1)-p(x)$.
- If $p$ is non-constant, the degree of $\delta p$ is the degree of $p$ minus one;
- If the leading term of $p(x)$ is $ax^n$ with $n\geq 1$, the leading term of $(\delta p)(x)$ is $na x^{n-1}$;
- If $p(x)=\binom{x}{k}$ with $k\geq 1$, then $(\delta p)(x)=\binom{x}{k-1}$;
- Any polynomial can be represented in the binomial base and with such representation the operator $\delta$ essentially acts as a shift by the previous point.
The given problem can be stated as $$ (\delta^2 P)(x) = 6\binom{x+1}{1} $$ hence a solution is given by $$ P(x) = 6\binom{x+1}{3} = \color{blue}{x^3-x}.$$ $(\delta^2 P)(x)$ equals zero iff the degree of $P$ is $\leq 1$, hence the full set of solutions is given by $\color{blue}{x^3+ax+b}$.
Let $\;P(x+1)-P(x)= Q(x)\;$ so $\;Q(x)-Q(x-1)=6x$
by induction, $Q(x+n)-Q(x)= 6((x+n)+(x+n-1)+...+(x+1))$
For $\,x=0\,$: $\displaystyle\;\;Q(n)=Q(0) + 6\big(n+(n-1)+\cdots+1\big)=Q(0)+3n(n+1)\,$, then:
$$\,P(n)=Q(n-1)+Q(n-2)+\cdots+Q(0)+P(0)= P(0)+n\,Q(0)+ 3\sum_{k=0}^{n-1} k(k+1)=\cdots$$
It should be $P(x)=ax^3+bx^2+cx+d$.
The condition gives $6ax+2b=6x$.
Thus, $a=1$, $b=0$ and $P(x)=x^3+cx+d$.
$$P(x)=P(0)+P'(0)x+\frac{P'(0)}{2!}x^2+...+\frac{P^{(n)}(0)}{n!}x^n=$$
$$=P(1)+P'(1)(x-1)+\frac{P'(1)}{2!}(x-1)^2+...+\frac{P^{(n)}(1)}{n!}(x-1)^n=$$ $$=P(-1)+P'(1)(x+1)+\frac{P'(-1)}{2!}(x+1)^2+...+\frac{P^{(n)}(-1)}{n!}(x+1)^n,$$ which gives $$P(x+1)=P(1)+P'(1)x+\frac{P'(1)}{2!}x^2+...+\frac{P^{(n)}(1)}{n!}x^n$$ and $$P(x-1)=P(-1)+P'(-1)x+\frac{P'(-1)}{2!}x^2+...+\frac{P^{(n)}(-1)}{n!}x^n.$$ Thus, $$6x=P(x+1)-2P(x)+P(x+1)=$$ $$=P(1)-2P(0)+P(-1)+\left(P'(1)-2P'(0)+P'(-1)\right)x+$$ $$+\frac{P''(1)-2P''(0)+P''(-1)}{2!}x^2+...\frac{P^{(n)}(1)-2P^{(n)}(0)+P^{(n)}(-1)}{n!}x^n$$ and since