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Mathematics
Problem
Let
S
k
=
∑
i
=
0
∞
1
(
k
+
1
)
i
S_k = \sum_{i=0}^{\infty} \frac{1}{(k+1)^i}
S
k
=
∑
i
=
0
∞
(
k
+
1
)
i
1
, then
∑
k
=
1
n
k
S
k
\sum_{k=1}^{n} kS_k
∑
k
=
1
n
k
S
k
equal : (a)
n
(
n
+
1
)
2
\frac{n(n+1)}{2}
2
n
(
n
+
1
)
(b)
n
(
n
−
1
)
2
\frac{n(n-1)}{2}
2
n
(
n
−
1
)
(c)
n
(
n
+
2
)
2
\frac{n(n+2)}{2}
2
n
(
n
+
2
)
(d)
n
(
n
+
3
)
2
\frac{n(n+3)}{2}
2
n
(
n
+
3
)
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