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(C) Given $n p(x) = (1 + x) p'(x)$.Let $p(x) = \sum_{k=0}^n a_k x^k$ with $a_n = 1$.The equation is $n \sum_{k=0}^n a_k x^k = (1 + x) \sum_{k=1}^n k a

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$b$ is a power of $2$

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(C) Given $n p(x) = (1 + x) p'(x)$.Let $p(x) = \sum_{k=0}^n a_k x^k$ with $a_n = 1$.The equation is $n \sum_{k=0}^n a_k x^k = (1 + x) \sum_{k=1}^n k a_k x^{k-1}$.Comparing the coefficients of $x^k$ on both sides:$n a_k = (k+1) a_{k+1} + k a_k$.This simplifies to $(n-k) a_k = (k+1) a_{k+1}$,or $a_{k+1} = \frac{n-k}{k+1} a_k$.Since $a_n = 1$,we have $a_{n-1} = \frac{n-(n-1)}{n-1+1} a_n = \frac{1}{n} 	imes n = \binom{n}{1}$.By induction,$a_{n-k} = \binom{n}{k}$.Thus,$p(x) = \sum_{k=0}^n \binom{n}{k} x^{n-k} = (x+1)^n$.Then $b = p(1) = (1+1)^n = 2^n$.Therefore,$b$ is a power of $2$.

Let $n > 2$ be an integer and define a polynomial $p(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$,where $a_0, a_1, \ldots, a_{n-1}$ are integers. Suppose we know that $n p(x) = (1 + x) p'(x)$. If $b = p(1)$,then:

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