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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 :-)

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    $\begingroup$ Don't close this question, no teacher would assign this $\endgroup$ – AmateurMathPirate Apr 1 '18 at 12:09
  • $\begingroup$ @AmateurMathGuy: This post lacks source and motivation entirely. Being homework is a sufficient reason to vote to close a question, but not a necessary one. $\endgroup$ – Carl Mummert Apr 1 '18 at 14:56
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    $\begingroup$ @CarlMummert Well, professor, it includes that he/ they are stuck. If source and motivation need to explicitly be displayed for every post, then why aren't they made into a field such that you can't post a question unless you fill this field? And every question has face value to a segment of users on MSE, perhaps there are some independent studiers who haven't even studied calculus yet, in this instance. In general, I have little indication that the virtue of just answering the question, or leaving off so someone else can benefit from it is observed by 3000 rep members of MSE $\endgroup$ – AmateurMathPirate Apr 1 '18 at 15:39
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$$P_n(x)=P_{n-1}(x-n)$$

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

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

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

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

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

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

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

coff . of x=

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

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    $\begingroup$ Observe that coeff. of $x$ is $P_{20}'(0)$. We have $P_{20}'(x)=3(x-210)^2+313\cdot 2(x-210)-77$ which gives us precisly what is written above. $\endgroup$ – szw1710 Apr 1 '18 at 12:16
  • $\begingroup$ Yours is the best answer as it used the binomial theorem and no calculus since the tags were algebra and precalculus $\endgroup$ – AmateurMathPirate Apr 1 '18 at 15:30
  • $\begingroup$ @ AmateurMathGuy: thank you so much bro.. $\endgroup$ – Faraday Pathak Apr 1 '18 at 16:34
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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 $$P'_0(a)-aP''(a)+a^2\frac{P'''(a)}2.$$

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