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CIIM
2012 CIIM
Problem 6
Problem 6
Part of
2012 CIIM
Problems
(1)
CIIM 2012 Problem 6
Source:
6/9/2016
Let
n
≥
2
n \geq 2
n
≥
2
and
p
(
x
)
=
x
n
+
a
n
−
1
x
n
−
1
+
⋯
+
a
1
x
+
a
0
p(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
p
(
x
)
=
x
n
+
a
n
−
1
x
n
−
1
+
⋯
+
a
1
x
+
a
0
a polynomial with real coefficients. Show that if there exists a positive integer
k
k
k
such that
(
x
−
1
)
k
+
1
(x-1)^{k+1}
(
x
−
1
)
k
+
1
divides
p
(
x
)
p(x)
p
(
x
)
then
∑
j
=
0
n
−
1
∣
a
j
∣
>
1
+
2
k
2
n
.
\sum_{j=0}^{n-1}|a_j| > 1 +\frac{2k^2}{n}.
j
=
0
∑
n
−
1
∣
a
j
∣
>
1
+
n
2
k
2
.
CIIM
CIIM 2012
undergraduate