Urgent!! If the sum of first n nos. of AP is cn2 then, the sum of the squares of these n terms is..
Sum of first n terms = Sn= cn2
So,Sum of first (n-1) terms = Sn-1 = c(n-1)2
n th term = tn = Sn- Sn-1 = c{n2 – (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 = c2(2n-1)2
Sum of n terms of the series of squares of the respective terms
= Sn’ = ∑c2(2n-1)2 = c2∑(2n-1)2 = c2∑(4n2-4n+1) = c2{4∑n2-4∑n+∑1}
= c2{4n(n+1)(2n+1)/6 – 4n(n+1)/2 + n}
= n(4n2-1)c2/3