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Putnam
1949 Putnam
B4
Putnam 1949 B4
Putnam 1949 B4
Source: Putnam 1949
March 20, 2022
Putnam
expansion
Problem Statement
Show that the coefficients
a
1
,
a
2
,
a
3
,
…
a_1 , a_2 , a_3 ,\ldots
a
1
,
a
2
,
a
3
,
…
in the expansion
1
4
(
1
+
x
−
1
1
−
6
x
+
x
2
)
=
a
1
x
+
a
2
x
2
+
a
3
x
3
+
…
\frac{1}{4}\left(1+x-\frac{1}{\sqrt{1-6x+x^{2}}}\right) =a_{1} x+ a_2 x^2 + a_3 x^3 +\ldots
4
1
(
1
+
x
−
1
−
6
x
+
x
2
1
)
=
a
1
x
+
a
2
x
2
+
a
3
x
3
+
…
are positive integers.
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