Progressions-12 (IIT-JEE 2009)

If the sum of first n terms of A.P is cn², then the sum of squares of these n terms is:

(a)n(4n²-1)c²/6          (b) n(4n²+1)c²/3        (c) n(4n²-1)c²/3        (d) n(4n²+1)c²/6

Solution follows here:

Solution:

Sum of first n terms = S_n= cn²

∴Sum of first (n-1) terms = S_n-1 = c(n-1)²

n th term = t_n = S_n- S_n-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  = t_n^’ = {c(2n-1)}² = c²(2n-1)²

Sum of n terms of the series of squares of the respective terms

= S_n^’ = ∑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

Answer (c)