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National and Regional Contests
India Contests
India IOQM
2020-21 IOQM India
3
\sum_{k=1}^{N} (2k+1)/(k^2+k)^2}= 0.9999 - IOQM 2020-21 p3
\sum_{k=1}^{N} (2k+1)/(k^2+k)^2}= 0.9999 - IOQM 2020-21 p3
Source:
January 18, 2021
Sum
algebra
Problem Statement
If
∑
k
=
1
N
2
k
+
1
(
k
2
+
k
)
2
=
0.9999
\sum_{k=1}^{N} \frac{2k+1}{(k^2+k)^2}= 0.9999
∑
k
=
1
N
(
k
2
+
k
)
2
2
k
+
1
=
0.9999
then determine the value of
N
N
N
.
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