if n
is a positive integer, show that,
(n+1)^2+(n+2)^2 +........+4n^2=n/6(2n+1)(7n+1)
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n
∑
n
=
1
n
2
=
n
6
(
n
+
1
)
(
2
n
+
1
)
Therefore,
2
n
∑
n
=
1
n
2
=
2
n
6
(
2
n
+
1
)
(
4
n
+
1
)
2
n
∑
n
+
1
n
2
=
2
n
∑
n
=
1
n
2
−
n
∑
n
=
1
n
2
=
2
n
6
(
2
n
+
1
)
(
4
n
+
1
)
−
n
6
(
n
+
1
)
(
2
n
+
1
)
=
n
6
(
2
(
2
n
+
1
)
(
4
n
+
1
)
−
(
n
+
1
)
(
2
n
+
1
)
)
=
n
6
(
(
16
n
2
+
12
n
+
2
)
−
(
2
n
2
+
3
n
+
1
)
)
=
n
6
(
14
n
2
+
9
n
+
1
)
=
n
6
(
7
n
+
1
)
(
2
n
+
1
)
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