​ If the sum of first ′n′ terms of an A.P. is cn^2, prove that sum of squares of these ‘n′ term is nc^2(4n^2-1)/3).

Question

If the sum of first ′n′ terms of an A.P. is cn², prove that sum of squares of these ‘n′ term is \(\frac{nc^2(4n^2−1)}{3}\).

Answer 1

Sum of first n terms = S_n = cn²                (given)

So, sum of first \((n − 1)\, terms = S_{n−1} = c(n − 1)^2\)

n^th term = \(t_n = S_n − S_{n−1}\)

= \(c\{n^2 − (n − 1)^2\}\)

= c(2n − 1)

To find sum of squares of fint n terms of fa series, we need n^th term of the series of squatesd the respective terms.

n^th term of the series of squares of the respective terms i.e.,

T_n = {c(2n − 1)}²

= c²(2n − 1)²

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

\(S_n = ∑C^2 (2n − 1)^2\)

= \(c^2∑(2n − 1)^2\)

= \(c^2∑(4n^2 − 4n + 1)\)

= \(c^2\bigg\{4∑n^2 − 4∑n + ∑1\bigg\}\)

= c² \(\bigg\{4 \bigg(\frac {n(n + 1)(2n + 1)}{6}\bigg)\) − 4 \(\bigg(\frac{n(n + 1)}{2} \bigg) + n\bigg\}\)

= nc²\(\bigg\{\frac{2(n + 1)(2n + 1)}{3} − 2(n + 1) + 1\bigg\}\)

= \(\frac{nc^2}{3} \){(n + 1)(2n + 1) − 6(n + 1) + 3}

= \(\frac{nc^2}{3} \)(4n² + 6n + 2 − 6n − 6 + 3)

= \(\frac{nc^2}{3} \)(4n² − 1)

Hence proved