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Mathematics
Mathematics
Problem
If
∑
r
=
1
n
T
r
=
n
(
n
+
1
)
(
n
+
2
)
3
\sum_{r=1}^{n} T_r = \frac{n(n+1)(n+2)}{3}
∑
r
=
1
n
T
r
=
3
n
(
n
+
1
)
(
n
+
2
)
, then
lim
n
→
∞
∑
r
=
1
n
2008
T
r
=
\lim_{n \to \infty} \sum_{r=1}^{n} \frac{2008}{T_r} =
lim
n
→
∞
∑
r
=
1
n
T
r
2008
=
(a) 2008 (b) 3012 (c) 4016 (d) 8032
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