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

1.	If the sum of first n terms of an A.P. is cn2, then the sum of the squares of these n terms is(a) \(\frac{n(4n^2-1)c^2}{6}\)(b) \(\frac{n(4n^2+1)c^2}{6}\)(c) \(\frac{n(4n^2-1)c^2}{3}\)(d) \(\frac{n(4n^2+1)c^2}{3}\)
Answer» (c) \(\frac{n(4n^2-1)c^2}{3}\) Let the sum of first n terms of the A.P, Sn = cn² ∴ S_{n – 1} = c (n – 1)² = c (n² – 2n + 1) ∴ nth term of the A.P. = S_n – S_{n – 1} = cn² – c (n² – 2n + 1) = c (2n – 1) Let \(S_n^2\) be the sum of the squares of these n terms. Then, \(S_n^2\) = \(\displaystyle\sum_{k=1}^{n} t_k{^2}\) where t_k = c (2k – 1) = \(\displaystyle\sum_{k=1}^{n} [c^2(2k-1)^2]\) = \(\displaystyle\sum_{k=1}^{n} [c^2.4k^2-c^2.4k+c^2]\) = \(4c^2\displaystyle\sum_{k=1}^{n} k^2\) - \(4c^2\displaystyle\sum_{k=1}^{n} k+c^2n\) = 4c²x \(\frac{n(n+1)(2n+1)}{6}\) - 4c² x \(\frac{n(n+1)}{2}\) + c²n = \(\frac{c^2n}{6}\)[4(c+1)(2n+1)-12(n+1)+6] = \(\frac{c^2n}{6}\)[8n² + 12n + 4 - 12n - 12 + 6] = \(\frac{c^2n}{6}\)[8n² - 2] = \(\frac{n(4n^2-1)c^2}{3}\)

If the sum of first n terms of an A.P. is cn2, then the sum of the squares of these n terms is(a) \(\frac{n(4n^2-1)c^2}{6}\)(b) \(\frac{n(4n^2+1)c^2}{6}\)(c) \(\frac{n(4n^2-1)c^2}{3}\)(d) \(\frac{n(4n^2+1)c^2}{3}\)

Answer»

(c) \(\frac{n(4n^2-1)c^2}{3}\)

Let the sum of first n terms of the A.P, Sn = cn²

∴ S_{n – 1} = c (n – 1)² = c (n² – 2n + 1)

∴ nth term of the A.P. = S_n – S_{n – 1}

= cn² – c (n² – 2n + 1) = c (2n – 1)

Let \(S_n^2\) be the sum of the squares of these n terms. Then,

\(S_n^2\) = \(\displaystyle\sum_{k=1}^{n} t_k{^2}\) where t_k = c (2k – 1)

= \(\displaystyle\sum_{k=1}^{n} [c^2(2k-1)^2]\) = \(\displaystyle\sum_{k=1}^{n} [c^2.4k^2-c^2.4k+c^2]\)

= \(4c^2\displaystyle\sum_{k=1}^{n} k^2\) - \(4c^2\displaystyle\sum_{k=1}^{n} k+c^2n\)

= 4c²x \(\frac{n(n+1)(2n+1)}{6}\) - 4c² x \(\frac{n(n+1)}{2}\) + c²n

= \(\frac{c^2n}{6}\)[4(c+1)(2n+1)-12(n+1)+6]

= \(\frac{c^2n}{6}\)[8n² + 12n + 4 - 12n - 12 + 6]

= \(\frac{c^2n}{6}\)[8n² - 2] = \(\frac{n(4n^2-1)c^2}{3}\)