If 1,S2,S3 are the sum of first n natural numbers, their squares and their cubes respectively, then
S1 =

n

Σ

k=1

k =

n(n+1)

2

S2 =

n

Σ

k=1

k2 =

n(n+1)(2n+1)

6

S3 =

n

Σ

k=1

k3 = [

n(n+1)

2

]2
Now, 1 + 8S1 = 1 + 8 [

n(n+1)

2

] = 1 + 4n (n + 1) = 4n2 + 4n + 1 = (2n+1)2
Now S3 (1 + 8 S1) = [

n(n+1)

2

]2(2n+1)2 = [

n(n+1)(2n+1)

2

]2
=

9

9

[

n(n+1)(2n+1)

2

]2 = 9[

n(n+1)(2n+1)

6

]2
= 9S22
Hence, 9S22 = S3(1+8S1)