# Question 7 & 8 Review Exercise 7

Solutions of Question 7 & 8 of Review Exercise 7 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.

Q7 For every positive intege: n. prove that $7^n-3^n$ is divisible by 4 . Solution: We using mathematical induction to prove the given statement. (1.) For $n=1$ then $7^k-3^k=7-4=4$. Thus 4 divides 4. Hence given is true for $n=1$. (2.) Let it be true for $n=k>1$ then $7^n-3^n=4 Q$ where $Q$ is the quotient in the induction hypothesis. (3.) For $n=k+1$ then we have \begin{aligned} & 7^{k+1}-3^{k+1}=7.7^k-3.3^k \\ & =(4+3) \cdot 7^k-3.3^k \\ & =4.7^k+3.7^k-3.3^k \end{aligned} \begin{aligned} & =4.7^k+3\left[7^k-3^k\right] \\ & \Rightarrow 7^{k+1}-3^{k+1}=4.7^k+3.4 Q \end{aligned} by induction hypothesis \begin{aligned} & \Rightarrow 7^{k=1}-3^{k+1}=4\left[7^k+3 Q\right] \\ & \Rightarrow 7^{k-1}-3^{k+1} \text { is divisible by } 4 . \end{aligned}

Thus the given statement is true for $n=k+1$. Hence it is true for all positive integers.

Q8 Prove that $(1+x)^n \geq(1+n x)$, for all natural number $n$ where $x>-1$. - Solution: We try to prove this using mathernatical induction. 1. For $n=1$ then $$(1+x)^1=1+x=1+1 x$$

Thus it is true for $n=1$. 2. Let it be true for $n=k$ then $$(1+x)^k \geq(1+k x)$$ 3. Now for $\boldsymbol{n}=k+1$ then we have \begin{aligned} & (1+x)^{k+1}=(1+x)^k(1+x) \\ & \geq(1+k x)(1+x) \text { by (a) } \\ & =1+x+k x+k x^2 \\ & =1+(k+1) x+k x^2 \\ & \geq 1+(k+1) x \\ & \Rightarrow(1+x)^{k+1} \geq[1+(k+1) x] . \end{aligned}

Hence the given is true for $n=k+1$. Thus by mathematical induction the given is true for all $n \geq 1$.