Chapter 12.9, Problem 19E

To determine

To prove: that the given proposition is valid for all positive integral values of $n$ by using mathematical induction.

Answer to Problem 19E

See explanation

Explanation of Solution

Given information:

  $5^n-2^n$ is divisible by 3, or it can be stated as follows:

  $5^n-2^n=3A$ for some integer A divisible by 3 (i.e. has a factor 3)

Calculation:

Let the statement be $S_n$ . Now verify that the statement is valid for $n=1$ .

  $5^1-2^1=3$

Since the factor is 3 so it is valid for $n=1$ .

Assume that the statement is valid for $n=k$ , i.e. $S_k\text{: }{5}^k-2^k=3A$ for some integer A divisible by 3 (i.e. has a factor 3).

Now verify that the statement is valid for $n=k+1$ , i.e. $S_{k+1}:5^{k+1}-2^{k+1}=3B$ for some integer B .

  $\begin{array}{c}{5}^k-2^k=3A\\ 5^k=2^k+3A\\ 5^{k+1}=\left(2^{k+1}+3A\right)\left(2+3\right)\\ =2^{k+1}+3\left(2^k\right)+6A+9A\\ 5^{k+1}-2^{k+1}=3\left(2^k\right)+15A\\ =3\left(2^k+5A\right)\\ 5^{k+1}-2^{k+1}=3B\end{array}$

Thus, if the given proposition is valid for $n=k$ then it is also valid for $n=k+1$ . Since the given proposition is valid $n=1$ then it is also valid for $n=2,n=3$ and so on. Thus, the given statement is valid for all integers.