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2014 CHMMC (Fall)
1
2014 CHMMC Tiebreaker 1 - sum a_i/(k^2+i)=1/k^2
2014 CHMMC Tiebreaker 1 - sum a_i/(k^2+i)=1/k^2
Source:
March 1, 2024
algebra
CHMMC
Problem Statement
For
a
1
,
.
.
.
,
a
5
∈
R
a_1,..., a_5 \in R
a
1
,
...
,
a
5
∈
R
,
a
1
k
2
+
1
+
.
.
.
+
a
5
k
2
+
5
=
1
k
2
\frac{a_1}{k^2 + 1}+ ... +\frac{a_5}{k^2 + 5}=\frac{1}{k^2}
k
2
+
1
a
1
+
...
+
k
2
+
5
a
5
=
k
2
1
for all
k
∈
{
2
,
3
,
4
,
5
,
6
}
k \in \{2, 3, 4, 5, 6\}
k
∈
{
2
,
3
,
4
,
5
,
6
}
. Calculate
a
1
2
+
.
.
.
+
a
5
6
.
\frac{a_1}{2}+... +\frac{a_5}{6}.
2
a
1
+
...
+
6
a
5
.
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