The cubes of the natural numbers are grouped as 1^3, (2^3, 3^3), (4^3,5^3,6^3),................ the sum of the numbers in the nth group is?
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Answer:
Series using the last element of the each group is
1
3
,3
3
,6
3
,10
3
,...
∴t
n
of 1,3,6,10,... is
2
1
n(n+1)
Now
if we replace n by n−1 we get the last element of (n−1)
th
group we add 1 in it and cubing the number that
number be the first element of n
th
group. So the
terms in the n
th
group are
(
2
n(n−1)
+1)
3
,(
2
n(n−1)
+2)
3
,...(
2
n(n+1)
)
3
∴ Sum of the element of n
th
group is
=sum of the cubes of first
2
n(n+1)
natural numbers − sum of the cubes of first
2
n(n−1)
natural numbers
=
4
(
2
n(n+1)
)
2
(
2
n(n+1)
+1)
2
−
4
(
2
n(n−1)
)
2
(
2
n(n−1)
+1)
2
=
8
n
3
[(n
2
+1)(n
3
+3)] (on simplification)
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