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Let $S$ denote the sum of the $2011th$ powers of the roots of the polynomial $(x − 2^0)(x − 2^1)\cdots(x − 2^{2010}) − 1$. How many $1$’s are in the binary expansion of $S$?

Progress: I used Newton's sums in order to find that

$a_nS_1 + a_{n−1} = 0$;

$a_nS_2 + a_{n−1}S_1 + 2a_{n-2} = 0$

$\hspace{20 mm}\vdots$

$a_nS_n +\cdots+a_1S_1 +na_0 =0$.

Then I was thinking of subtracting $1$ from $a_0$. After that, the sum of the $2011th$ roots is now $1+2^{2011} +2^{4022}+\cdots+2^{2010\cdot2011} +2011$?

user19405892
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1 Answers1

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We have two polynomials, $P(x) = (x-2^0)(x-2^1)\cdots(x-2^{2010})$ and $T(x)=(x-2^0)(x-2^1)\cdots(x-2^{2010}) -1 $. The goal is to relate those two equations to each other. We do so as I said, but when subtracting $1$ we must increase $S_n$ by $\dfrac{n}{a_n} = 2011$. Therefore, the answer is simply the sum of the $2011th$ roots of $P(x)$ plus $\dfrac{n}{a_n}$, which is $1+2^{2011} +2^{4022} +· · ·+2^{2010·2011} + 2011$. The task remains to convert this into binary, which is done as follows: $2011_2 = 11111011100_2$ and the sum of the remaining terms is $1000000\cdots1000000\cdots $ (where there are $2010$ ones in binary). So the digit sum is $2010 + 8 = 2018$.

user19405892
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