﻿ Step by Step derivation for the sum of the N<sup>2</sup>

## Step by step derivation for the Sum of n2 and the Sum of higher integer power.

Question : How to derive the formula for the sum of n2, which is stated as
n k=1 k2 =
(n)(n + 1)(2n+1) / 6
?

Solution : We make use of the fact that both equation (1) and (2) as shown below are true.

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]

Thus,

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

= n3 + 3
[
(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]

By applying this criteria, We can get the sum of ANY higher integer power, eg.
n k=1 k3
,
n k=1 k4
,
n k=1 k5
... ...

Below is a working example to get
n k=1 k3 =
(n)2(n + 1)2 / 4

Again, we make use of the fact that both equation (3) and (4) as shown below are true.

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)

Thus,

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)

#### 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]

[back to solution]

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