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Poland - Second Round
2014 Poland - Second Round
3.
Smallest possible value
Smallest possible value
Source: Poland 2014 - Second Round P3
July 28, 2019
algebra
polynomial
Polynomials
Problem Statement
For each positive integer
n
n
n
, determine the smallest possible value of the polynomial
W
n
(
x
)
=
x
2
n
+
2
x
2
n
−
1
+
3
x
2
n
−
2
+
…
+
(
2
n
−
1
)
x
2
+
2
n
x
.
W_n(x)=x^{2n}+2x^{2n-1}+3x^{2n-2}+\ldots + (2n-1)x^2+2nx.
W
n
(
x
)
=
x
2
n
+
2
x
2
n
−
1
+
3
x
2
n
−
2
+
…
+
(
2
n
−
1
)
x
2
+
2
n
x
.
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