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2002
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Day 1, Problem 3
IMC 2002 · Problem 3
Day 1
9th IMC · Warsaw, Poland
Statement
Let
n
n
n
be a positive integer and let
a
k
=
1
(
n
k
)
,
b
k
=
2
k
−
n
,
for
k
=
1
,
2
,
…
,
n
.
a_k = \frac{1}{\binom{n}{k}}, \quad b_k = 2^{k-n}, \quad \textit{for} \quad k = 1, 2, \dots, n.
a
k
=
(
k
n
)
1
,
b
k
=
2
k
−
n
,
for
k
=
1
,
2
,
…
,
n
.
Show that
a
1
−
b
1
1
+
a
2
−
b
2
2
+
⋯
+
a
n
−
b
n
n
=
0.
(1)
\frac{a_1 - b_1}{1} + \frac{a_2 - b_2}{2} + \dots + \frac{a_n - b_n}{n} = 0. \tag{1}
1
a
1
−
b
1
+
2
a
2
−
b
2
+
⋯
+
n
a
n
−
b
n
=
0.
(
1
)
Official solution
Reveal
Hidden so you can work on the problem first.
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