(i) nth term of the given series

Tn = 1² + 2² + 3² + ..... + n² = ∑n² = \(\frac{n(n+1)(2n+1)}{6}\) = \(\frac{2n^3+3n^2+n}{6}\) = \(\frac{n^3}{3}\) + \(\frac{n^2}{2}\) + \(\frac{n}{6}\)

∴ S_n = \(\frac{1}{3}\). ∑n³ + \(\frac{1}{2}\).∑n² + \(\frac{1}{6}\). ∑n = \(\frac{1}{3}\).\(\frac{n^2(n+1)^2}{4}\) + \(\frac{1}{2}\).\(\frac{n(n+1)(2n+1)}{6}\) + \(\frac{1}{6}\).\(\frac{n(n+1)}{4}\)

= \(\frac{n(n+1)}{12}\) {n(n + 1) + (2n + 1) + 1} 

= \(\frac{n(n+1)(n^2+3n+2)}{12}\) = \(\frac{n(n+1)(n+1)(n+2)}{12}\) 

= \(\frac{n^2(n+1)^2(n+2)}{12}\)

(ii) In the given series each term is the product of two factors. The factors 3, 4, 5, ..... are in A.P. having 3 as the first term and 1 as the common difference, therefore the nth term of this A.P. = 3 + (n – 1)1 = 2 + n. Also, the factors 5, 7, 9, .... are in A.P. having 5 as the first term and 2 as the common difference, therefore the nth term of this A.P. = 5 + (n – 1)² = 2n + 3

∴ nth term of the series = (2 + n) (2n + 3) 

⇒ T_n = 2n² + 7n + 6

∴ S_n = 2 . ∑n² + 7∑n + 6n = \(\frac{2n(n+1)(2n+1)}{6}\) + \(\frac{7n(n+1)}{2}\) + 6n

= \(\frac{2n(n+1)(2n+1)+21n(n+1)+36n}{6}\) = \(\frac{4n^3+27n^2+59n}{6}\)