Question

1.2
If the sum of first n terms of an A.P. is cn2,then the sum of squares of these n terms is
[IIT-JEE - 2009, Paper-2, (3, -1), 80)
n(4n? - 1) c?
Bin(4n? +1) c?
in n(4n² - 1) ca
no ve son paper 212 -1,
(4n2
(D) n(4n? +1) C2
(A)

Answer 1

Answer:

Step-by-step explanation:

Given  

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

Given Sum of first n terms Sn = cn^2

 So sum of first (n – 1) terms = Sn-1 = c(n – 1)^2

Now n th term = tn = Sn – Sn-1

                             = cn^2 – c(n – 1)^2

                             = c(n^2 – (n – 1)^2)

                              = c(2n – 1)

Now n th term of series of squares of terms will be  

                    tn = {c(2n – 1)}^2

                      = c^2(2n – 1)^2

Now sum of n terms of series of squares of terms is given by

Sn = ∑ c^2(2n – 1)^2

    = c^2 ∑ 4n^2 – 4n + 1)

    = c^2 4 ∑ n^2 – 4 ∑ n + ∑ 1

    = c^2 4n (n + 1)(2n + 1) / 6 – 4n(n + 1) / 2 + n

    = nc^2 (4n^2 – 1) / 3