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Putnam
2013 Putnam
3
Putnam 2013 A3
Putnam 2013 A3
Source:
December 9, 2013
Putnam
algebra
polynomial
function
integration
college contests
Problem Statement
Suppose that the real numbers
a
0
,
a
1
,
…
,
a
n
a_0,a_1,\dots,a_n
a
0
,
a
1
,
…
,
a
n
and
x
,
x,
x
,
with
0
<
x
<
1
,
0<x<1,
0
<
x
<
1
,
satisfy
a
0
1
−
x
+
a
1
1
−
x
2
+
⋯
+
a
n
1
−
x
n
+
1
=
0.
\frac{a_0}{1-x}+\frac{a_1}{1-x^2}+\cdots+\frac{a_n}{1-x^{n+1}}=0.
1
−
x
a
0
+
1
−
x
2
a
1
+
⋯
+
1
−
x
n
+
1
a
n
=
0.
Prove that there exists a real number
y
y
y
with
0
<
y
<
1
0<y<1
0
<
y
<
1
such that
a
0
+
a
1
y
+
⋯
+
a
n
y
n
=
0.
a_0+a_1y+\cdots+a_ny^n=0.
a
0
+
a
1
y
+
⋯
+
a
n
y
n
=
0.
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