(i) 1.3 + 3.5 + 5.7 + …
nth term of the given series
Tn = (1 + 3 + 5 + … nth term)
(3 + 5 + 7 + … nth term)
= [1 + (n – 1)2 [3 +(n – 1)2]
= (1 + 2n – 2) (3 + 2n – 2)
= (2n – 1) (2n + 1)
∴ Sn = ∑ (4n2 – 1)
= 4∑n2 – ∑1
(ii) 1.2.4 + 2.3.7 + 3.4.10 + …
nth term of the given series
Tn = (1 + 2 + 3 + … nth term)
(2 + 3 + 4 + … nth term )
(4 + 7 + 10 + … nth term)
= n . (n + 1) . [4 + (n – 1)3]
= n(n + 1) (3n + 1)
= (n2 + n) (3n + 1)
= 3n3 + 3n2 + n2 + n
= 3n3 + 4n2 + n = n(n + 1) (3n + 1)
∴ Sn = ∑(3n3 + 4n2 + n)
= 3∑n3 + 4∑n2 + ∑M
