On oberservation,
(i) a=1,2,3,4...
Here, we have a pattern of (n+1) where n is a whole number.
By substituting n=0,1,2,3..,
a=0+1,1+1,2+1,3+1...
=1,2,3,4... ⇒n+1
(ii) b=0,2,4,6,8...
Here, we have pattern of 2n where n is a whole number.
By substituting n=0,1,2,3..,
b=2×0,2×1,2×2,2×3...
=0,2,4,6... ⇒2n