Chapter 9.5, Problem 8E

Solution

To state : an explicit rule for the $n^{th}$ term of recursively defined sequence $a_n=5a_{n-1},a_1=3$ and then use mathematical induction to prove this.

An explicit rule for the $n^{th}$ term of recursively defined sequence $a_n=3a_{n-1},a_1=2$ is $a_n=2\cdot 3^{n-1}$ .

Given information :

The recursively defined sequence $a_n=5a_{n-1},a_1=3$

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 $a_n=5a_{n-1},a_1=3$ ,

  $a_1=3$

  $\begin{array}{c}{a}_2=5a_1\\ =5^1\cdot 3\end{array}$

  $\begin{array}{c}{a}_3=5a_2\\ =5\cdot 5\cdot 3\\ =5^2\cdot 3\end{array}$

  $\begin{array}{c}{a}_4=5a_3\\ =5\cdot 5^2\cdot 3\\ =5^3\cdot 3\end{array}$

And so on.

The sequence is a geometric sequence with $a_1=3$ and $r=5$ .

So, the general term is calculated using the formula $a_n=a{r}^{n-1}$ .

  $a_n=3\cdot 5^{n-1}$

Now, to prove that $a_n=3\cdot 5^{n-1}$ true holds for all integers using mathematical induction.

Conjecture:

Take $P_n$ to be statement $a_n=3\cdot 5^{n-1}$ .

Anchor step:

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

For $n=1$ ,

  $\begin{array}{c}{a}_1=3\\ 3\cdot 5^{1-1}=3\\ 3\cdot 5^0=3\\ 3=3\end{array}$

Inductive hypothesis:

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

  $P_k:a_k=3\cdot 5^{k-1}$

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

  $a_k=3\cdot 5^{k-1}$

Replace $a_k$ with $3\cdot 5^{k-1}$ as $P_k$ is assumed to true.

  $\begin{array}{c}{a}_{k+1}=5\cdot 3\cdot 5^{k-1}\\ =3\cdot 5^k\end{array}$

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, $P_n:a_n=3\cdot 5^{n-1}$ is true for all integers $n$ .