Step by Step derivation for the sum of the N<sup>2</sup>
Question : How to derive the formula for the sum of n2, which is stated as
Solution : We make use of the fact that both equation (1) and (2) as shown below are true.
Thus,
= n3 +
3
By applying this criteria, We can get the sum of ANY higher integer power, eg.
Below is a working example to get
Again, we make use of the fact that both equation (3) and (4) as shown below are true.
Thus,
=
=
=
[back to solution]
Step by step derivation for the Sum of n2 and the Sum of higher integer power.
n
∑
k=1 k2 =
?
(n)(n + 1)(2n+1)
6
n
∑
k=1 k3 -
n
∑
k=1 (k -1)3 = n3
. . . . . . . . . . . . . . . . . . . . .(1) [proof]
n
∑
k=1 k3 -
n
∑
k=1 (k -1)3 =
n
∑
k=1 (3k2 -3k +1)
. . . . . . . . . . .(2) [proof]
n
∑
k=1 (3k2 -3k +1) = n3
3
n
∑
k=1 k2 -
3
n
∑
k=1 k +
n
∑
k=1 1 = n3
3
n
∑
k=1 k2 = n3 +
3
n
∑
k=1 k -
n
∑
k=1 1
[
(n)(n + 1)
2
] - n
n
∑
k=1 k2
=
2n3 +3n(n + 1) -2n
6
=
n(2n2 +3n+ 1)
6
=
n(n +1)(2n+ 1)
6
[Shown]
n
∑
k=1 k3
,
n
∑
k=1 k4
,
n
∑
k=1 k5
... ...
n
∑
k=1 k3 =
(n)2(n + 1)2
4
n
∑
k=1 k4 -
n
∑
k=1 (k -1)4 = n4
. . . . . . . . . . . . . . . . . . . . .(3)
n
∑
k=1 k3 -
n
∑
k=1 (k -1)3 =
n
∑
k=1 (4k3 -6k2 +4k -1)
. . . . . . . . . . .(4)
n
∑
k=1 (4k3 -6k2 +4k -1)
= n4
4
n
∑
k=1 k3 -
6
n
∑
k=1 k2 +
4
n
∑
k=1 k -
n
∑
k=1 1 = n4
4
n
∑
k=1 k3 = n4+
6
n
∑
k=1 k2 -
4
n
∑
k=1 k -
n
∑
k=1 1
n4
+6[
(n)(n + 1)(2n + 1)
6
]-
4[
(n)(n + 1)
2
] +n
n
∑
k=1 k3
=
n4+2n3 + 3n2 + n -2n2 -2n +n
4
n4+2n3 + n2
4
=
n2(2n2 +2n +1)2
4
(n)2(n + 1)2
4
(Shown)
Appendix
Proof for equation(1):
n
∑
k=1 k3 -
n
∑
k=1 (k -1)3
= [(13 +23 +33 +43 +... ... +(n-1)3 +n3] - [(13 +23 +33 +43 +... ... +(n-1)3]
=n3
[proven]
Proof for equation(2):
n
∑
k=1 k3 -
n
∑
k=1 (k -1)3 =
n
∑
k=1 k3 -(k3 -3k2 +3k -1)
=
n
∑
k=1 (3k2 -3k +1)
[proven]