====== 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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