# Question 13 Exercise 7.1

Solutions of Question 13 of Exercise 7.1 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

$2^n>n \forall n \in \mathbf{N}$.

1. For $n=1$ then $2^n=2^1=2$ and $n=1$.

Clearly $2>1$. hence the given statement is true for $n=1$.

2. Let it be true for $n=l>I$ then $2^k>k\cdots(i)$

3. For $n=k+1$ then we consider \begin{align} & 2^{k+1}=2^k \cdot 2>k \cdot 2 \quad \text { by (i) } \\ & \Rightarrow 2^{k+1}>2 k=k+k \\ &\Rightarrow 2^{k+1}>k+1 \text {. as } k>1\end{align} Which is the form of proposition taken when $n$ is replace by $k+1$,

hence true for $n=k+1$. Thus by mathematical induction it is true for all $n \in \mathbf{N}$.

$n$ ! $>n^2$ for every integer $n \geq 4$

1. For $n=4$ then $n !=4 !=24$ and $n^2=4^2=16$.

Clearly $24>16$, hence the given proposition is true for $n=4$.

2. Let it be true for $n=k>4$ then $k !>k^2\cdots(i)$

3. For $n=k+1$ then we have \begin{align} & (k+1) !=(k+1) k !>(k+1)(k+1) \\ & \because k !>k+1 \\ & \Rightarrow(k+1) !>(k+1)^2 . \end{align} Which is the form taken by proposition when $n$ is replaced by $k+1$,

hence it is true for $n=k+1$. Thus by mathematical induction it is true for all $n \geq 4$.