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6
Induction with divisibility
Induction with divisibility
Source: Moldova TST 2020
March 8, 2020
Divisibility
algebra
Problem Statement
Let
n
n
n
,
(
n
≥
3
)
(n \geq3)
(
n
≥
3
)
be a positive integer and the polynomial
f
(
x
)
=
(
1
+
x
)
⋅
(
1
+
2
x
)
⋅
(
1
+
3
x
)
⋅
.
.
.
⋅
(
1
+
n
x
)
f(x)=(1+x) \cdot (1+2x) \cdot (1+3x) \cdot ... \cdot (1+nx)
f
(
x
)
=
(
1
+
x
)
⋅
(
1
+
2
x
)
⋅
(
1
+
3
x
)
⋅
...
⋅
(
1
+
n
x
)
=
a
0
+
a
1
⋅
x
+
a
2
⋅
x
2
+
a
3
⋅
x
3
+
.
.
.
+
a
n
⋅
x
n
= a_0+a_1 \cdot x+a_2 \cdot x^2+a_3 \cdot x^3+...+a_n \cdot x^n
=
a
0
+
a
1
⋅
x
+
a
2
⋅
x
2
+
a
3
⋅
x
3
+
...
+
a
n
⋅
x
n
. Show that the number
a
3
a_3
a
3
divides the number
k
=
C
n
+
1
2
⋅
(
2
⋅
C
n
2
⋅
C
n
+
1
2
−
3
⋅
a
2
)
.
k=C^2_{n+1} \cdot (2 \cdot C^2_n \cdot C^2_{n+1}-3 \cdot a_2).
k
=
C
n
+
1
2
⋅
(
2
⋅
C
n
2
⋅
C
n
+
1
2
−
3
⋅
a
2
)
.
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