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China Contests
China National Olympiad
2005 China National Olympiad
4
fractional sequence inequality
fractional sequence inequality
Source: china mathematical olympiad cmo 2005 final round - Problem 4
January 25, 2005
inequalities
calculus
function
derivative
logarithms
algebra unsolved
algebra
Problem Statement
The sequence
{
a
n
}
\{a_n\}
{
a
n
}
is defined by:
a
1
=
21
16
a_1=\frac{21}{16}
a
1
=
16
21
, and for
n
≥
2
n\ge2
n
≥
2
,
2
a
n
−
3
a
n
−
1
=
3
2
n
+
1
.
2a_n-3a_{n-1}=\frac{3}{2^{n+1}}.
2
a
n
−
3
a
n
−
1
=
2
n
+
1
3
.
Let
m
m
m
be an integer with
m
≥
2
m\ge2
m
≥
2
. Prove that: for
n
≤
m
n\le m
n
≤
m
, we have
(
a
n
+
3
2
n
+
3
)
1
m
(
m
−
(
2
3
)
n
(
m
−
1
)
m
)
<
m
2
−
1
m
−
n
+
1
.
\left(a_n+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left(m-\left(\frac{2}{3}\right)^{{\frac{n(m-1)}{m}}}\right)<\frac{m^2-1}{m-n+1}.
(
a
n
+
2
n
+
3
3
)
m
1
m
−
(
3
2
)
m
n
(
m
−
1
)
<
m
−
n
+
1
m
2
−
1
.
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