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

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

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

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

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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## Comments

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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 \[2016*2017=\boxed{4066272}\] – Aareyan Manzoor · 1 year, 3 months ago

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– Otto Bretscher · 1 year, 3 months ago

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.Log in to reply

– Aareyan Manzoor · 1 year, 3 months ago

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

– Otto Bretscher · 1 year, 3 months ago

So when you add up \(\sum_{i}\) you have 2017 terms.Log in to reply

– Aareyan Manzoor · 1 year, 3 months ago

i have said why we can neglect -1, reducing a root and giving us 2016 terms.Log in to reply

– Otto Bretscher · 1 year, 3 months ago

But if you use only 2016 terms then it is no longer true that the sum of the 2016th to first powers is zero. Think about it... you will see what I mean.Log in to reply

– Aareyan Manzoor · 1 year, 3 months ago

true... the answer should be \(2017^2\)?Log in to reply

– Otto Bretscher · 1 year, 3 months ago

No, there is another small error that compensates for this one... you are losing a term \(-1\) along the way... the constant term in your sum should be 2017-1=2016, so it all works out... a Christmas miracle ;)Log in to reply

– Aareyan Manzoor · 1 year, 3 months ago

sorry for late response( electricity...), what a miracle indeed, i have edited it.Log in to reply

– Otto Bretscher · 1 year, 3 months ago

One more small correction: In your double sum, it is \(i\) that goes to 2017, not \(k\).Log in to reply

– Aareyan Manzoor · 1 year, 3 months ago

corrected once again...Log in to reply

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 \(j\leq 2016\). – Otto Bretscher · 1 year, 3 months ago

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– Chinmay Sangawadekar · 1 year, 3 months ago

@Otto Bretscher , please try to solve using Vieta.Log in to reply

– Otto Bretscher · 1 year, 3 months ago

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)Log in to reply

@Otto Bretscher, please view my comment when you have time. – Aareyan Manzoor · 1 year, 3 months ago

yes,you are rightLog in to reply

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. – Otto Bretscher · 1 year, 3 months ago

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– Aareyan Manzoor · 1 year, 3 months ago

hmm... i wrote that feature in my wiki "roots of unity" and proved it by the linked note. the same thing really.Log in to reply

– Chinmay Sangawadekar · 1 year, 3 months ago

This also would workLog in to reply

@Calvin Lin , what is your opinion for this ? What will be your approach ? – Chinmay Sangawadekar · 1 year, 3 months ago

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@Otto Bretscher ,

Cool problem .... the new problem which is based on our yesterday's conversation ... – Chinmay Sangawadekar · 1 year, 3 months ago

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– Otto Bretscher · 1 year, 3 months ago

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!Log in to reply

@Otto Bretscher – Chinmay Sangawadekar · 1 year, 3 months ago

Bonus problem is complicated thoughLog in to reply

– Otto Bretscher · 1 year, 3 months ago

No it's not! Not for a guy like you ;)Log in to reply

@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 – Chinmay Sangawadekar · 1 year, 3 months ago

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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 ? – Chinmay Sangawadekar · 1 year, 3 months ago

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– Aareyan Manzoor · 1 year, 3 months ago

nope.Log in to reply

@Aareyan Manzoor – Chinmay Sangawadekar · 1 year, 3 months ago

So am I right or wrong ?Log in to reply

@Aareyan Manzoor , but why ? – Chinmay Sangawadekar · 1 year, 3 months ago

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– Aareyan Manzoor · 1 year, 3 months ago

i dont get "we seek a polynomial ...."Log in to reply

@Aareyan Manzoor – Chinmay Sangawadekar · 1 year, 3 months ago

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

– Aareyan Manzoor · 1 year, 3 months ago

we really didnt use vietas. otto sir used roots of unity, i used a version of newtons sum.Log in to reply

@Aareyan Manzoor – Chinmay Sangawadekar · 1 year, 3 months ago

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

– Aareyan Manzoor · 1 year, 3 months ago

i think not....Log in to reply

– Otto Bretscher · 1 year, 3 months ago

Aareyan, are you not using Viete to make sure that the symmetric sums of the \(x_i\) are zero?Log in to reply

@Aareyan Manzoor , @Otto Bretscher – Chinmay Sangawadekar · 1 year, 3 months ago

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

– Otto Bretscher · 1 year, 3 months ago

In my approach, I'm not using Viete and I certainly don't need such a polynomial.Log in to reply

@Otto Bretscher – Chinmay Sangawadekar · 1 year, 3 months ago

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 ? – Chinmay Sangawadekar · 1 year, 3 months ago

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– Otto Bretscher · 1 year, 3 months ago

Yes, Aareyan is applying Viete to the polynomial \(x^{2017}+1\)... that works.Log in to reply