Chapter 9.5, Problem 6E

Solution

To state : an explicit rule for the $n^{th}$ term of recursively defined sequence $a_n=a_{n-1}+2,a_1=7$ and then use mathematical induction to prove this.

An explicit rule for the $n^{th}$ term of recursively defined sequence $a_n=a_{n-1}+2,a_1=7$ is $a_n=2n+5$ .

Given information :

The recursively defined sequence $a_n=a_{n-1}+2,a_1=7$

Formula used :

Follow the steps to prove that the statement is true for any positive integer.

Anchor step: Prove that any statement $P_n$ is true for $n=1$ .

Inductive hypothesis: Assume $P_n$ is true for $n=k$ .

Inductive Step: Prove that $P_n$ is true for $n=k+1$ .

Explanation :

Consider the recursively defined sequence $a_n=a_{n-1}+2,a_1=7$ ,

The sequence is an arithmetic sequence with $a_1=7$ and $d=2$ .

So, the general term is calculated using the formula $a_n=a+\left(n-1\right)d$ .

  $\begin{array}{c}{a}_n=7+2\left(n-1\right)\\ =7+2n-2\\ a_n=2n+5\end{array}$

Now, to prove that $a_n=2n+5$ true holds for all integers using mathematical induction.

Conjecture:

Take $P_n$ to be statement $a_n=2n+5$ .

Anchor step:

Prove that $P_n$ is true for $n=1$ .

For $n=1$ ,

  $\begin{array}{c}{a}_1=7\\ 2\cdot 1+5=7\\ 2+5=7\\ 7=7\end{array}$

Inductive hypothesis:

Assume $P_n$ is true for $n=k$ . Replace $n$ with $k$ in $P_n$ .

  $P_k:a_k=2k+5$

Inductive Step:

Prove that $P_n$ is true for $n=k+1$ . This means it is needed to be proved that $P_{k+1}$ must be true.

Proceed with the sequence formula

  $a_{k+1}=a_k+2$

Replace $a_k$ with $2k+5$ as $P_k$ is assumed to true.

  $\begin{array}{c}{a}_{k+1}=2k+5+2\\ =\left(2k+2\right)+5\\ =2\left(k+1\right)+5\end{array}$

The statement is same as the statement $P_{k+1}$ , so $P_{k+1}$ is true.

So, if $P_n$ is true for $n=k$ , then it is also true for $n=k+1$ .

Conclusion:

As $P_n$ is true for $n=1$ and $P_k$ further implies $P_{k+1}$ , this means $P_n$ is true for $n=2$ , $n=3$ and so on.

By mathematical induction, $P_n:a_n=5n-2$ is true for all integers $n$ .