Set inclusion relationship

Note $6^{2n} = 36^n = (1 + 35)^n$. By Binomial Theorem: $36^n - 35n - 1 = (1 + 35n + \binom{n}{2}35^2 + \dots) - 35n - 1 = 35^2 \left(\binom{n}{2} + \dots\right) = 1225k$ So every element in $A$ is a multiple of 1225.

Set $B = \{35(n-1) : n = 1,2,3,\dots\} = \{0, 35, 70, 105, \dots\}$, which is the set of all non-negative multiples of 35.

Since 1225 is a multiple of 35, every element in $A$ is a multiple of 35, so $A \subseteq B$. However, $35 \in B$ but $35 \notin A$ (since smallest non-zero element in $A$ is $1225$). Thus, every member of A is in B and at least one member of B is not in A.