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2021 LMT Spring
A17
A17
Part of
2021 LMT Spring
Problems
(1)
2021 LMT Spring Division A Problem 17
Source:
10/22/2021
Given that the value of
∑
k
=
1
2021
1
1
2
+
2
2
+
3
2
+
⋯
+
k
2
+
∑
k
=
1
1010
6
2
k
2
−
k
+
∑
k
=
1011
2021
24
2
k
+
1
\sum_{k=1}^{2021} \frac{1}{1^2+2^2+3^2+\cdots+k^2}+\sum_{k=1}^{1010} \frac{6}{2k^2-k}+\sum_{k=1011}^{2021} \frac{24}{2k+1}
k
=
1
∑
2021
1
2
+
2
2
+
3
2
+
⋯
+
k
2
1
+
k
=
1
∑
1010
2
k
2
−
k
6
+
k
=
1011
∑
2021
2
k
+
1
24
can be expressed as
m
n
\frac{m}{n}
n
m
, where
m
m
m
and
n
n
n
are relatively prime positive integers, find
m
+
n
m+n
m
+
n
.Proposed by Aidan Duncan