Question

If $A=\left\{5^{n}-4 n-1: n \in N\right\}$ and $B=\{16(n-1): n \in N\}$, then
(A) $A=B$
(B) $A \cap B=\phi$
(C) $A \subseteq B$
(D) $B \subseteq A$

Answer 1

Ans: (C)
$5^{n}-4 n-1=(4+1)^{n}-4 n-1=16 k, k \in Z, A$ is a set of some multiple of 16 while set $B$ is the set of all consecutive multiple of $16 .$

Answer 2

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Answer 3

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