Prove that number of subsets of a set containing $n$ distinct elements is $2^n$, for all $n \in N$.
Let $P(n)$ : Number of subset of a set containing $n$ distinct elements is $2^n$, for all $n \in N$.
Step I We observe that $P(1)$ is true, for $n=1$.
Number of subsets of a set contain 1 element is $2^1=2$, which is true.
Step II Assume that $P(n)$ is true for $n=k$.
$P(k)$ : Number of subsets of a set containing $k$ distinct elements is $2^k$, which is true.
Step III To prove $P(k+1)$ is true, we have to show that
$P(k+1)$ : Number of subsets of a set containing $(k+1)$ distinct elements is $2^{k+1}$.
We know that, with the addition of one element in the set, the number of subsets become double.
$\therefore$ Number of subsets of a set containing $(k+1)$ distinct elements $=2 \times 2^k=2^{k+1}$. So, $P(k+1)$ is true. Hence, $P(n)$ is true.
If $10^n+3 \cdot 4^{n+2}+k$ is divisible by 9 , for all $n \in N$, then the least positive integral value of $k$ is
For all $n \in N, 3 \cdot 5^{2 n+1}+2^{3 n+1}$ is divisible by
If $x^n-1$ is divisible by $x-k$, then the least positive integral value of $k$ is
If $P(n): 2 n < n!, n \in N$, then $P(n)$ is true for all $n \geq$ ........... .
Given that, $$P(n):2n < n!,n \in N$$
$$\begin{aligned} \text{For n = 1, }\quad 2 & < ! \quad \text{[false] }\\ \text{For n = 2, }\quad 2 \times 2< 2!4 & < 2 \quad \text{[false] }\\ \text{For n = 3, }\quad 2 \times 3 & < 3! \\ 6 & < 3! \\ 6 & < 3 \times 2 \times 1 \\ (6 & < 6) \quad \text{[false] }\\ \text{For n = 4, }\quad 2 \times 4 & < 4! \\ 8 & < 4 \times 3 \times 2 \times 1 \\ (8 & < 24) \quad \text{[true] }\\ \text{For n = 5, }\quad 2 \times 5 & < 5! \\ 10 & < 5 \times 4 \times 3 \times 2 \times 1 \\ (10 & < 120) \quad \text{[true] } \end{aligned}$$
Hence, $P(n)$ is for all $n\ge4$.