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If
∑
n
=
1
100
tan
−
1
(
2
n
2
+
n
+
4
)
=
tan
−
1
(
p
q
)
\sum_{n=1}^{100} \tan^{-1} \left( \frac{2}{n^2 + n + 4} \right) = \tan^{-1} \left( \frac{p}{q} \right)
∑
n
=
1
100
tan
−
1
(
n
2
+
n
+
4
2
)
=
tan
−
1
(
q
p
)
…
Mathematics
Problem
If
∑
n
=
1
100
tan
−
1
(
2
n
2
+
n
+
4
)
=
tan
−
1
(
p
q
)
\sum_{n=1}^{100} \tan^{-1} \left( \frac{2}{n^2 + n + 4} \right) = \tan^{-1} \left( \frac{p}{q} \right)
∑
n
=
1
100
tan
−
1
(
n
2
+
n
+
4
2
)
=
tan
−
1
(
q
p
)
,
p
,
q
∈
N
p, q \in \mathbb{N}
p
,
q
∈
N
, then least value of
(
p
−
q
)
(p - q)
(
p
−
q
)
is
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