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If $A=\left\{5^{n}-4 n-1: n \in N\right\}$ and
$B=\{16(n-1): n \in N],$ then
Options:
$B=\{16(n-1): n \in N],$ then
Solution:
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Verified Answer
The correct answer is:
$A \subseteq B$
We have,
$\begin{aligned} A &=5^{n}-4 n-1=(1+4)^{n}-4 n-1 \\ &=\left({ }^{n} C_{0}+{ }^{n} C_{1} \times 4+{ }^{n} C_{2} \times 4^{2}+\ldots+{ }^{n} C_{n} \times 4^{n}\right) \\ &=4^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} \times 4+\ldots+{ }^{n} C_{n} \times 4^{n-2}\right) \end{aligned}$
$\therefore$ A contains some multiples of 16
Clearly, $B$ contains all multiples of 16 including $0 .$
$A \subseteq B$
$\begin{aligned} A &=5^{n}-4 n-1=(1+4)^{n}-4 n-1 \\ &=\left({ }^{n} C_{0}+{ }^{n} C_{1} \times 4+{ }^{n} C_{2} \times 4^{2}+\ldots+{ }^{n} C_{n} \times 4^{n}\right) \\ &=4^{2}\left({ }^{n} C_{2}+{ }^{n} C_{3} \times 4+\ldots+{ }^{n} C_{n} \times 4^{n-2}\right) \end{aligned}$
$\therefore$ A contains some multiples of 16
Clearly, $B$ contains all multiples of 16 including $0 .$
$A \subseteq B$
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