Question on coefficients of polynomials

Question

Let $$P_0(x)=x^3+313x^2-77x-8$$

For integers $n\ge1$, define $$P_n(x)=P_{n-1}(x-n)$$.

Then what is the coefficient of $x$ in $P_{20}(x)$ ?

I have absolutely no idea how to proceed. Could somebody help?

Thanks for any help :-)

Answer 1

$$P_n(x)=P_{n-1}(x-n)$$

$$ P_{20}(x)=P_{19}(x-20)$$

$$ = P_{18}(x-20-19)=P_{17}(x-20-19-18)$$

$$ = ...=P_{0}(x-210)$$

$$ P_{20} (x)=(x-210)^3+313(x-210)^2-77(x-210)-8 $$

The coefficient of x in the above expression is its derivative $$ 3(x-210)^2+626(x-210)-77 $$ at $x=0$, which is $763$

Answer 2

$P_{20} (x)=\left(x-\displaystyle\sum_{r=1}^{r=20} r \right)^3+313\left(x-\displaystyle\sum_{r=1}^{r=20} r \right)^2 -77\left(x-\displaystyle\sum_{r=1}^{r=20} r \right)-8$

$P_{20} (x)=(x-210)^3+313(x-210)^2-77(x-210)-8$

coff . of x=

$ 3\cdot(210)^2+313\cdot(-420)-77=763$

Answer 3

Hint:

You can easily find a formula for $P_n(x)$ $$P_n(x)=P_0\biggl(x-\frac{n(n+1)}2\biggr).$$ Next, use Taylor's formula for polynomials. As $P_0$ has degree $3$, $$P_0(x-a)=P_0(a)+P'_0(a)(x-a)+\frac{P''(a)}2(x-a)^2+\frac{P'''(a)}6(x-a)^3.$$ Thus we obtain the coefficient of $x$ is $$P'_0(a)-aP''(a)+a^2\frac{P'''(a)}2.$$