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Mathematics
Problem
If
∑
r
=
1
n
T
r
=
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
64
\sum_{r=1}^{n} T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}
∑
r
=
1
n
T
r
=
64
(
2
n
−
1
)
(
2
n
+
1
)
(
2
n
+
3
)
(
2
n
+
5
)
, then
lim
n
→
∞
∑
r
=
1
n
(
1
T
r
)
\lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{T_r} \right)
lim
n
→
∞
∑
r
=
1
n
(
T
r
1
)
is equal to : (1) 1 (2) 0 (3)
2
3
\frac{2}{3}
3
2
(4)
1
3
\frac{1}{3}
3
1
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