MA017 2010 HOMEWORK 4
(1)
Show that
n
X
k
=0
(

1)
k
(
n
k
)
(
k
+ 1)
2
=
1
n
+ 1
n
+1
X
k
=1
1
k
.
Consider (1

x
)
n
=
∑
(

1)
k
(
n
k
)
. Integrate both sides, then divide both sides by
x
and integrate
again (be careful with the boundary terms).
(2)
Prove that all terms of the sequence
a
1
=
a
2
=
a
3
= 1
,
a
n
+1
=
1 +
a
n

1
a
n
a
n

2
are integers.
This was a Putnam problem, but I don’t remember the year. You can solve it by
looking for a more symmetric form recursion, which leads naturally to the discovery of an invariant,
as follows. Looking at the given recursion for
n
and
n
+ 1:
a
n
+1
a
n

2
= 1 +
a
n

1
a
n
,
1 +
a
n
a
n
+1
=
a
n
+2
a
n

1
.
Adding these two equations, we get
a
n
+1
(
a
n

2
+
a
n
) =
a
n

1
(
a
n
+
a
n
+2
)
,
i.e. that
a
n

2
+
a
n
a
n

1
=
a
n
+
a
n
+2
a
n
+1
.
Thus
b
n
:=
a
n
+
a
n
+2
a
n
+1
satisﬁes the recursion
b
n
+2
=
b
n
. We have
a
1
= 1
,a
2
= 1
,a
3
= 1
,a
4
= 2, so that
b
1
= 2 and
b
2
= 3;
thus in general
b
2
n
= 1
,b
2
n
+1
= 3. The recursion
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 Fall '09
 symmetric form recursion, m,n a1 a1

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