Q64 / 155 · WBJEE 2016 · Mathematics

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
    $A \cap B=\phi$
  • Correct answer
    C
    $A \subseteq B$
  • D
    $B \subseteq A$

Solution

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 $0 .$ $A \subseteq B$

Hence, the correct answer is option: 'C'.

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