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1995 AIME Problems
13
Summation
Summation
Source:
December 22, 2005
floor function
inequalities
Problem Statement
Let
f
(
n
)
f(n)
f
(
n
)
be the integer closest to
n
4
.
\sqrt[4]{n}.
4
n
.
Find
∑
k
=
1
1995
1
f
(
k
)
.
\sum_{k=1}^{1995}\frac 1{f(k)}.
∑
k
=
1
1995
f
(
k
)
1
.
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