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We know that every $d$-degree polynomial can be uniquely expressed as $P(n)=\sum_{k=0}^da_k\binom{n}{k}$ so that $a_k=\sum_{i=0}^k(-1)^{k-i}\binom{k}{i}P(i)$. In particular $\sum_{i=0}^d(-1)^{d-i}\binom{d}{i}P(i)=c$, where $c$ is the leading coefficient of $P(n)$.

Now, introduce $Q(n)=P(a+bn)$. Find $\sum_{i=0}^d(-1)^{d-i}\binom{d}{i}Q(i)$ in terms of $c,a,b,d$.

From the answers, I have $c\cdot d!\cdot b^d$. And I don't see how they are getting it... $Q(n)$ is a $d$-degree polynomial with a leading coefficient of $c\cdot b^d$. Shouldn't the sum equal $c\cdot b^d$ then, and not $c\cdot d!\cdot b^d$? Where do they get $d!$ from?

Thanks.

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Let $Q(n)=b_d\binom{n}{d}+\dots+b_0\binom{n}{0}$, then: $$ b_d=\sum_{i=0}^d(-1)^{d-i}\binom{d}{i}Q(i)$$

So this where the confusion gets in, $b_d$ is not the leading coefficient of polynomial $Q$ in the classical sense, as in the term by the coefficient $n^d$.

Then the term by $n^d$ in expression $Q(n)$ is $\frac{b_d}{d!}$ as $b_d\binom{n}{d}=b_d\cdot\frac{n(n-1)\dots(n-d+1)}{d!}$. This is equal to $c\cdot b^d$ if $c$ is the leading coefficient of $P(n)$ in the classical sense, so that $b_d=c\cdot b^d\cdot d!$.

But if $c$ is as you wrote, then it's a wrong result. Then $\frac{b_d}{d!}$ must be equal to $\frac{c\cdot b^d}{d!}$ so your answer is right.

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  • $\begingroup$ Lol, I totally forgot about how factorials in the denominators of $\binom{n}{k}$ affect the coefficients. Thanks. $\endgroup$ – user75619 Jul 8 '18 at 11:21

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