# Question 14 Exercise 7.1

Solutions of Question 14 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.

## Question 14 (i)

Show that $5$ is a factor $3^{2 n-1}+2^{2 n-1}$ where $n$ is any positive integer.

### Solution

1. For $n=1$ then $$3^{2 n-1}+2^{2 n-1}=3^{2.1-1}+2^{2.1-1}=5 \text {. }$$ $5$ divides $5$, hence $5$ is a factor of $5.$ Thus given is true for $n=1$

2. Let it be true for $n=k>1$,

then $54$ divides $3^{2 k} 1+2^{2 k} \quad 1$ which implies that $$3^{2 k-1}+2^{2 k-1}=5 Q$$ where $Q$ is a quotient.

3. For $n=k+1$ then considering \begin{align} 3^{2(k+1)-1}+2^{2(k+1)-1} & =3^{2 k+2-1}+2^{2 k+2-1}\\ & =3^{2 k} \cdot 3^2+2^{2 k-1} \cdot 2^2 \\ & =3^{2 k-1} \cdot(5+4)+2^{2 k-1} \cdot 4 \\ & =5.3^{2 k}+4(3^{2 k-1}+2^{2 k-1}) \\ & =5.3^{2 k}+5.4 . Q\end{align} by induction hypouresis $$=5 [3^{2 k-1}+4 \cdot Q]$$ $5$ is a factor of $5[3^{2 h-1}+4()^{\prime \prime}$

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

## Question 14(ii)

Prove that $2^{2 n}-1$ is a multiple of $3$ for all natural numbers.

### Solution

1. For $n=1$ then $$2^{2 n}-1=2^{2.1}-1=4-1=3 $$ $3$ divides $3$, hence 3 is multiple of $3.$

Thus the statement is true for $n=1$.

2. Let it be true for $n=k$ then $3$ divides $2^{2 k}-1$ or $2^{2 k}-1=3 Q$

where $Q \subseteq \mathbb{Z}$ is quotient.

3. For $n=k+1$ then we have \begin{align} 2^{2(k+1)}-1&=2^{2 k+2}-1 \\ & =2^{2 k} \cdot 2^2-1 \\ & =2^{2 k}(3+1)-1 \\ & =3.2^{2 k}+2^{2 k}-1 \\ & =3.2^{2 k}+3 Q\end{align} by induction hypothesis $$=3[2^{2 k}+Q]$$ $3$ divides $3[2^{2 k}+Q]$ or $3[2^{2 k}+Q]$ is a multiple of $3.$

Hence the given statement is true for $n=k+1$.

Thus by mathematical induction the given statement is true for all $n \in \mathbf{N}$.

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