1000 Followers problem!

Algebra Level 5

α η α 2 α η = 0 \large \alpha^{\eta} - \alpha^2-\alpha-\eta=0

Consider the above polynomial and let its roots be β ( 1 , η ) , β ( 2 , η ) , β ( 3 , η ) , , β ( η , η ) \large\beta_{(1,\eta)},\beta_{(2,\eta)},\beta_{(3,\eta)} , \ldots, \beta_{(\eta,\eta)} .

Define S η = i = 1 η β ( i , η ) η \large S_{\eta}= \displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)}^{\eta} , find the value of j = 5 1000 S j \large \displaystyle\sum_{j=5}^{1000} S_j .


The answer is 333833470.

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2 solutions

Nihar Mahajan
Jun 5, 2015

If we have polynomial x 3 + a x 2 + b x + c x^3+ax^2+bx+c we denote σ 1 = p + q + r \sigma_1=p+q+r and σ 2 = p q + q r + p r \sigma_2=pq+qr+pr where p , q , r p,q,r are its roots.

If you consider all polynomials of kind α η α 2 α η = 0 \large\alpha^{\eta}-\alpha^2-\alpha-\eta=0 ,where 5 η 1000 \large5\leq\eta\leq 1000 , we notice that : σ 1 = σ 2 = 0 \large\boxed{\sigma_1=\sigma_2=0} .

Since β ( 1 , η ) \beta_{(1,\eta)} is a root of the given polynomial , we have :

β ( 1 , η ) η β ( 1 , η ) 2 β ( 1 , η ) η = 0 β ( 1 , η ) η = β ( 1 , η ) 2 + β ( 1 , η ) + η \Large{\beta_{(1,\eta)}^{\eta} - \beta_{(1,\eta)}^2 - \beta_{(1,\eta)} -\eta = 0 \\ \Rightarrow\beta_{(1,\eta)}^{\eta} = \beta_{(1,\eta)}^2 + \beta_{(1,\eta)} + \eta}

Similarly we have for β ( 2 , η ) , β ( 3 , η ) β ( η , η ) \large\beta_{(2,\eta)},\beta_{(3,\eta)} \dots \beta_{(\eta,\eta)} .Adding them all we get :

i = 1 η β ( i , η ) η = i = 1 η β ( i , η ) 2 + i = 1 η β ( i , η ) + η 2 i = 1 η β ( i , η ) η = σ 1 2 2 σ 2 + σ 1 + η 2 i = 1 η β ( i , η ) η = η 2 = S η \Large{\displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)}^{\eta} = \displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)}^2 + \displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)} + \eta^2 \\ \Rightarrow \displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)}^{\eta} = \sigma_1^2 - 2\sigma_2 + \sigma_1 + \eta^2 \\ \Rightarrow \displaystyle\sum_{i=1}^{\eta} \beta_{(i,\eta)}^{\eta} = \eta^2 = S_{\eta}}

Since 5 η 1000 5\leq\eta\leq 1000 , we have :

j = 5 1000 S j = (sum of squares from 1 to 1000) (sum of squares form 1 to 4) j = 5 1000 S j = ( 1000 ) ( 1001 ) ( 2001 ) 6 ( 4 ) ( 5 ) ( 9 ) 6 j = 5 1000 S j = 333833500 30 = 333833470 \displaystyle\sum_{j=5}^{1000} S_j = \text{(sum of squares from 1 to 1000)} - \text{(sum of squares form 1 to 4)} \\ \large{\Rightarrow \displaystyle\sum_{j=5}^{1000} S_j = \dfrac{(1000)(1001)(2001)}{6} - \dfrac{(4)(5)(9)}{6} \\ \Rightarrow \displaystyle\sum_{j=5}^{1000} S_j = 333833500 - 30 = \boxed{333833470}}

Congratulations on 1000 followers, @Nihar Mahajan .(Sorry, I am wishing you very late!)

Anuj Shikarkhane - 6 years ago

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Thanks very much. Congrats to you too :P

Nihar Mahajan - 6 years ago

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Thanks... :)

Anuj Shikarkhane - 6 years ago

Great problem, and nice solution.

Daniel Liu - 6 years ago

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Thanks!!! :)

Nihar Mahajan - 6 years ago

great problem but found difficult to understand

Sukrut Waghmare - 6 years ago

Nice problem. To me it was very scary at the beginning but as I get thinking about it , it showed to be much more easy than I thought at first sight.

I did the same path you wrote in your solution.

I just can't understand the first line of your solution. I think that if you write a polynomial x 3 + a x 2 + b x + c x^3 + ax^2 + bx +c then you should call a -a the sum of its roots. I think you meant to write the polynomial ( x a ) ( x b ) ( x c ) (x-a)(x-b)(x-c) . Am I right?

This is maybe what prevent some reader to understand the solution.

Andrea Palma - 6 years ago

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No , I was telling what σ 1 , σ 2 \sigma_1 , \sigma_2 means in that statement.

Nihar Mahajan - 6 years ago

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Yes, I was just telling that the letters a , b , c a,b,c have already been used as the coefficients of x 2 , x 1 , x 0 x^2,x^1,x^0 (respectively) in the polynomial you introduced.

Andrea Palma - 6 years ago

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@Andrea Palma Oh , that was my mistake sorry.Lemme edit it.

Nihar Mahajan - 6 years ago

What's the need of making this big the answer? It doesn't make sense. Questions with such big answers become uninteresting.

Kartik Sharma - 6 years ago

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According to me , the content/idea/theme in the problem makes it interesting. If there are tedious calculations or calculation complexities , you can use calculator as an aid. Since its a 1000 followers problem , I thought of including 1000 1000 in the problem.

I apologize for making the answer big , but the words " It doesn't make sense" and "uninteresting" hurt me because I took many efforts in creating this problem.I am extremely sorry for the trouble. I want all people to enjoy this problem.

Nihar Mahajan - 6 years ago

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Oh! I didn't mean to hurt you. I was just sharing my views. Using a calculator is an option but as I have said, you could have made it different to ease calculations. These big numbers in the answer box don't look good.

Kartik Sharma - 6 years ago

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@Kartik Sharma Okay , I will consider your opinion when I will post a question.Thanks!

Nihar Mahajan - 6 years ago
Rajdeep Dhingra
Jun 5, 2015

Let the η = y \eta = y and α = x \alpha = x .
We know the fact that
a 0 S 1 + a 1 = 0 a 0 S 2 + a 1 S 1 + 2 a 2 = 0 a 0 S r + a 1 S r 1 + . . . . . . + a r 1 S 1 + r a r = 0 a_0 S_1 + a_1 = 0 \\ a_0 S_2 + a_1 S_1 + 2a_2 = 0 \\ a_0S_r + a_1S_{r-1} + ...... + a_{r-1}S_1 + ra_r = 0 .

Now the Polynomial is P ( x ) = x y x 2 x y = 0 a 0 = 1 , a 1 = a 2 = . . . = a y 3 = 0 , a y 2 = 1 , a y 1 = 1 , a y = y P(x) = x^y - x^2 - x - y = 0 \\ \Rightarrow a_0 = 1 , a_1 = a_2 = ... = a_{y-3} = 0 , a_{y-2} = -1 , a_{y-1} = -1 , a_y = -y

So if we apply the above results we'll get S 1 = S 2 = S 3 = . . . = S y 3 = 0 S_1 = S_2 = S_3 = ... = S_{y-3} = 0 .
Further applying the result we'll get S y 2 = y 2 , S y 1 = y 1 , S y = y 2 S_{y-2} = y-2 , S_{y-1} = y-1 , S_{y} = y^2 .

So we now have to evaluate y = 5 1000 S y \displaystyle \sum_{y=5}^{1000} {S_y} . Which we know is equal to y = 5 1000 y 2 = 333833470 \displaystyle \sum_{y=5}^{1000}{y^2} = \boxed{333833470}

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