$\large \alpha^{\eta} - \alpha^2-\alpha-\eta=0$

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

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

The answer is 333833470.

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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

Great problem, and nice solution.

Daniel Liu
- 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 $x^3 + ax^2 + bx +c$ then you should call $-a$ the sum of its roots. I think you meant to write the polynomial $(x-a)(x-b)(x-c)$ . 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 $\sigma_1 , \sigma_2$ means in that statement.

Nihar Mahajan
- 6 years ago

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Yes, I was just telling that the letters $a,b,c$ have already been used as the coefficients of $x^2,x^1,x^0$ (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 $1000$ 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

Let the
$\eta = y$
and
$\alpha = x$
.

We know the fact that

$a_0 S_1 + a_1 = 0 \\ a_0 S_2 + a_1 S_1 + 2a_2 = 0 \\ a_0S_r + a_1S_{r-1} + ...... + a_{r-1}S_1 + ra_r = 0$
.

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

So if we apply the above results we'll get
$S_1 = S_2 = S_3 = ... = S_{y-3} = 0$
.

Further applying the result we'll get
$S_{y-2} = y-2 , S_{y-1} = y-1 , S_{y} = y^2$
.

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

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

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

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

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

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

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

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

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