\[ x^{2016}-x^{2015}+x^{2014}-\cdots+x^{2} -x+1\ \]

has roots , \(x_{1},x_{2},\ldots,x_{2016}\). Evaluate

\[\left ( \displaystyle \sum_{k=1}^{2016}\sum_{i=1}^{2016}(1+x_{i})^{k} \right )\]

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TopNewest-1 is obviously not a root. we multiply by (x+1) to find \[f(x)=x^{2017}+1=0\] one root of this expression is -1, and if we put it in the sum,i.e. \((1+x_i)=(1-1)=0\). so we can go on withput worries. we are just searching for: \[\sum_{k=1}^{2016}\sum_{i=1}^{2017} (x_i+1)^k=\sum_{i=1}^{2017} (\dfrac{(x_i+1)^{2017}-1}{x_i+1-1}-1)=\sum_{i=1} (x_i^{2016}+\binom{2017}{1}x_i^{2015}+....+\binom{2017}{2016}-1)\] gp used here. view this.yes, the sum of all 2016th to 1st power is zero. the \(\binom{2017}{2016}-1=2016 \)is left. added 2017 times results in \[2016*2017=\boxed{4066272}\]

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There may be a small error in the count (or rather two errors that cancel out). Your sum actually involves 2017 \(i\)'s (you have added -1) but you have to subtract 1 from the summand.

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i have said that "one root of this expression is -1...". at the first.

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Taking a quick look at this during our x-mas festivities, I would suspect that the inner sum is always 2017, so that the whole sum would be 2016*2017.

[added later] We observe that the \(y_i=-x_i\), together with \(y_0=1\), are the 2017th roots of unity. Now

\[\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-y_i)^k=\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=\sum_{k=1}^{2016}2017=2016*2017\]

since the sum of the \(j\)th powers of the 2017th roots of unity is 0 for \(j\leq 2016\).

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@Otto Bretscher , please try to solve using Vieta.

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My solution is quite short and straightforward.... why should I look for another solution? ;) (I guess I could use Viète to prove that the sum of the \(j\)th powers of the roots of unity is 0)

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yes,you are right @Otto Bretscher, please view my comment when you have time.

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I thought about it in a slightly different but equivalent way.

We observe that the \(y_i=-x_i\), together with \(y_0=1\), are the 2017th roots of unity. Now \(\sum_{i=0}^{2016}(1-y_i)^k\)\(=\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=2017\) since the sum of the \(j\)th powers of the 2017th roots of unity is 0.

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@Calvin Lin , what is your opinion for this ? What will be your approach ?

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@Otto Bretscher ,

Cool problem .... the new problem which is based on our yesterday's conversation ...

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Yes, thanks for that suggestion! For square-free numbers the count turns out to be pretty easy... the general case is a mess, though! Nobody has solved it yet... I'm counting on you!

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Bonus problem is complicated though @Otto Bretscher

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@Otto Bretscher , do we need to find a polynomial whose roots are , \(2017 , 2017^2 ,.... 2017^2016\) ? As the inner summation is 2017 @Aareyan Manzoor help

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@Otto Bretscher , @Aareyan Manzoor , I have slight doubt ,

the inner summation will be definitely 2017.

So we need a polynomial whose roots are , \(2017^{1} , 2017^{2} , 2017^{3} , .... , 2017^{2016}\) , isnt it ?

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nope.

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So am I right or wrong ? @Aareyan Manzoor

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@Aareyan Manzoor , but why ?

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@Aareyan Manzoor

sorry we need a polynomial , so that we can find the summation using vieta ,Log in to reply

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@Aareyan Manzoor

But can this be solved by vieta ? according to me it could notLog in to reply

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@Aareyan Manzoor , @Otto Bretscher

Lets try again write a solution on the work that you have done , on vieta approach , then I will post mineLog in to reply

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@Otto Bretscher

I am just asking whether my approach is right ? (by vieta)Log in to reply

@Otto Bretscher , but please also see the vieta approach ,,, is it right ?

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