it the sum of first n terms of A.P is cn^2,then the sum of squares of these n terms is
(1)n(4n^2-1)/c^2/6 (2)n(4n^2+1)/c^2/3 (3)n(4n^2-1)/c^2/3 (4)n(4n^3+1)/c^2/6

• Sum of first n terms = Sn= cn^2
  
  
  Sum of first (n-1) terms = Sn-1 = c(n-1)^2
  n th term = tn = Sn- Sn-1 = c{n^2 – (n-1)^2} = c(2n-1)
  To find sum of squares of first n terms of this series, we need n th
  term of the series of squares of the respective terms.
  n th term of the series of squares of the respective terms = tn
  ={c(2n-1)}^2 = c^2(2n-1)^2
  Sum of n terms of the series of squares of the respective terms
  = Sn
  = Σc2(2n-1)^2 = c^2Σ(2n-1)^2 = c^2Σ(4n^2-4n+1)
  =c^2{4Σn^2-4Σn+Σ1}
  = c^2{4n(n+1)(2n+1)/6 – 4n(n+1)/2 + n}
  = n(4n^2-1)c^2/3
• 11 years agoHelpfull: Yes(3) No(2)