α η − α 2 − α − η = 0
Consider the above polynomial and let its roots be β ( 1 , η ) , β ( 2 , η ) , β ( 3 , η ) , … , β ( η , η ) .
Define S η = i = 1 ∑ η β ( i , η ) η , find the value of j = 5 ∑ 1 0 0 0 S j .
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Congratulations on 1000 followers, @Nihar Mahajan .(Sorry, I am wishing you very late!)
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Thanks very much. Congrats to you too :P
Great problem, and nice solution.
great problem but found difficult to understand
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 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.
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No , I was telling what σ 1 , σ 2 means in that statement.
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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.
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@Andrea Palma – Oh , that was my mistake sorry.Lemme edit it.
What's the need of making this big the answer? It doesn't make sense. Questions with such big answers become uninteresting.
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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 1 0 0 0 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.
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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.
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@Kartik Sharma – Okay , I will consider your opinion when I will post a question.Thanks!
Let the
η
=
y
and
α
=
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
.
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
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 y = 5 ∑ 1 0 0 0 S y . Which we know is equal to y = 5 ∑ 1 0 0 0 y 2 = 3 3 3 8 3 3 4 7 0
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If we have polynomial x 3 + a x 2 + b x + c we denote σ 1 = p + q + r and σ 2 = p q + q r + p r where p , q , r are its roots.
If you consider all polynomials of kind α η − α 2 − α − η = 0 ,where 5 ≤ η ≤ 1 0 0 0 , we notice that : σ 1 = σ 2 = 0 .
Since β ( 1 , η ) 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
Similarly we have for β ( 2 , η ) , β ( 3 , η ) … β ( η , η ) .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
Since 5 ≤ η ≤ 1 0 0 0 , we have :
j = 5 ∑ 1 0 0 0 S j = (sum of squares from 1 to 1000) − (sum of squares form 1 to 4) ⇒ j = 5 ∑ 1 0 0 0 S j = 6 ( 1 0 0 0 ) ( 1 0 0 1 ) ( 2 0 0 1 ) − 6 ( 4 ) ( 5 ) ( 9 ) ⇒ j = 5 ∑ 1 0 0 0 S j = 3 3 3 8 3 3 5 0 0 − 3 0 = 3 3 3 8 3 3 4 7 0