Find the sum of the following series up to n terms: 

6 + .66 + .666+….

Given:

S₁ = n(n + 1) / 2

S₃ = n²(n + 1)² / 4

Now,

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

= [n²(n + 1)² / 4][1 + 4n² + 4n]

= [n²(n + 1)² / 4](2n + 1)²

= [n(n + 1)(2n + 1)]² / 4     ...(1)

Also,

9S₂² = 9[n(n + 1)(2n + 1)/6]²

= (9/36)[n(n + 1)(2n + 1)]²

= [n(n + 1)(2n + 1)]² / 4     ...(2)

Therefore, from (1) and (2):

9S₂² = S₃(1 + 8S₁)

Hence proved.