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Putnam
1972 Putnam
B1
Putnam 1972 B1
Putnam 1972 B1
Source: Putnam 1972
February 17, 2022
Putnam
Sequences
Problem Statement
Let
∑
n
=
0
∞
x
n
(
x
−
1
)
2
n
n
!
=
∑
n
=
0
∞
a
n
x
n
\sum_{n=0}^{\infty} \frac{x^n (x-1)^{2n}}{n!}=\sum_{n=0}^{\infty} a_{n}x^{n}
∑
n
=
0
∞
n
!
x
n
(
x
−
1
)
2
n
=
∑
n
=
0
∞
a
n
x
n
. Show that no three consecutive
a
n
a_n
a
n
can be equal to
0
0
0
.
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