**i. 3 + 7 + 13 + 21 + .......**

1st consecutive differences 4, 6, 8,........are in A.P.

∴ T_{n} = an^{2} + bn + c

Putting n = 1,2,3,

a + b + c = 3, 4a + 2b + c = 7, 9a + 3b + c = 13

⇒ a = 1, b = 1, c = 1

∴ T_{n} = n^{2} + n + 1

**ii. 5 + 7 + 13 + 31 + 85 +..........**

1st consecutive differences 2, 6, 18, 54,....... are in G.P. with common ratio 3

∴ T_{n} = a.3^{n–1} + bn + c

Putting n = 1,2,3 we get

a + b + c = 5, 3a + 2b + c = 7, 9a + 3b + c = 13

⇒ a = 1, b = 0, c = 4

∴ T_{n} = 3^{n–1} + 4