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Mathematics
Problem
For a positive integer
n
n
n
, the sum
∑
k
=
1
n
k
(
k
+
1
)
\displaystyle\sum_{k=1}^{n}k(k+1)
k
=
1
∑
n
k
(
k
+
1
)
is equal to a)
n
(
n
+
1
)
(
2
n
+
1
)
6
\displaystyle\frac{n(n+1)(2n+1)}{6}
6
n
(
n
+
1
)
(
2
n
+
1
)
b)
n
2
(
n
+
1
)
3
\displaystyle\frac{n^{2}(n+1)}{3}
3
n
2
(
n
+
1
)
c)
n
(
n
+
1
)
(
n
+
2
)
3
\displaystyle\frac{n(n+1)(n+2)}{3}
3
n
(
n
+
1
)
(
n
+
2
)
d)
n
(
n
+
1
)
(
3
n
+
1
)
6
\displaystyle\frac{n(n+1)(3n+1)}{6}
6
n
(
n
+
1
)
(
3
n
+
1
)
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