Question

Prove that 3 is a factor of ${\mathbf{2}}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$for every positive integer n.

Short answer

3 is a factor of $2^{2\mathrm{n}+1}$for every positive integern.

Step-by-step solution

Step 1

Consider the following expression,

$P\left(n\right)=2^{2n+1}+1$

Step 2

Substitute 1 for n in the equation$P\left(n\right)=2^{2n+1}+1$ as follows;

$$\begin{gathered} P\left(1\right)=2^{2\left(1\right)+1}+1 \\ =2^3+1 \\ =9 \end{gathered}$$

Since, 9 is divisible by 3 this implies that at n =1 , it is true.

Substitute 2 for n in the equation$P\left(n\right)=2^{2n+1}+1$ as follows;

$$\begin{gathered} P\left(2\right)=2^{2\left(1\right)+1}+1 \\ =2^5+1 \\ =33 \end{gathered}$$

Since, 33 is divisible by 3 this implies that at n = 2 , it is true.

Step 3

Substitute k for n in the equation $P\left(n\right)=2^{2n+1}+1$ as follows;

$P\left(k\right)=2^{2k+1}+1$

Suppose thatis divisible by 3m where m is an integer.

So,

$$\begin{gathered} 2^{2k+1}+1=3m \\ {2}^{2k+1}=3m-1 \end{gathered}$$

Step 4

Substitute k+1 for n in the equation $P\left(n\right)=2^{2n+1}+1$as follows;

$$\begin{gathered} P\left(k+1\right)=2^{2\left(k+1\right)+1}+1 \\ =2^{2k+2+1}+1 \\ =2^{2k+1}.2^2+1 \end{gathered}$$

Substitute$2^{2k+1}=3m-1$for in the equation $P\left(k+1\right)=2^{2k+1}.2^2+1$as follows;

$$\begin{gathered} P\left(k+1\right)=\left(3m-1\right).2^2+1 \\ =3m.2^2-1.2^2+1 \\ =3m.2^2-4+1 \\ =3\left(4m-1\right) \end{gathered}$$

Hence, $P\left(k+1\right)$ is divisible by 3 if P (k) is divisible by 3.