Chapter 8, Problem 1RQ

a) What is a recurrence re1aon?

b) Find a recurrence relation for the amount of money that will be in an account after n years if $1,000,000 is deposited in an account yielding 9% annually.

To determine

(a)

To define:

What recurrence relation is.

Answer to Problem 1RQ

A recurrence relation for a sequence $a_{0,}{a}_{1,\mathrm{..............}}{a}_{n-1.}$ is formula that relates each term $a_n$ to certain of its predecessors $a_{0,}{a}_1\mathrm{............}{a}_{n-1.}$

Explanation of Solution

A recurrence relation for a sequence $a_{0,}{a}_{1,\mathrm{..............}}{a}_{n-1.}$ is formula that relates each term $a_n$ to certain of its predecessors $a_{0,}{a}_1\mathrm{............}{a}_{n-1.}$

For example,

$\begin{array}{l}{a}_n=2a_{n-1}+1\\ r\ge 1\\ a_0=0\end{array}$

To determine

(b)

To find:

recurrence relation.

Answer to Problem 1RQ

$a_n=a_{n-1}+0.09a_{n-1}$

Explanation of Solution

Given:

$P=1,000,000\$,N=nyears,R=9\%$

Formula used:

$I=\frac{PNR}{100}$

Calculation:

here $P=1,000,000\$,N=nyears,R=9\%$ We suppose the amount will be in the amount in the account after n years be $a_n$.

Then this amount will be the amount in the interest after $n-1$ years plus interest on the account after $n-1$ years be $a_{n-1}$

Then interest at $9\%$ per annum will be

$\begin{array}{l}\frac{a_{n-1}\times 9\times 1}{100}\\ =0.09a_{n-1}\end{array}$

So $a_n=a_{n-1}+0.09a_{n-1}$

Initial $\$1,000,000$ is deposited in the account the initial condition is $a_0=1,000,000$