See Explanation below
Explanation:
a) The given Recursive formula:
[tex]\begin{gathered} u_n=0.75u_{n-1}\text{ + 210} \\ \text{where u}_n\text{ = 450} \end{gathered}[/tex]To get u1, n = 1
[tex]\begin{gathered} u_1=0.75u_{1-1}+\text{ 210} \\ u_1=0.75u_0\text{ + 210} \\ u_1=\text{ 0.75}\times450\text{ + 210} \\ u_1\text{ = 547.5} \end{gathered}[/tex]To get u2, n = 2
[tex]\begin{gathered} u_2=0.75u_{2-1}\text{ + 210} \\ u_2=0.75u_1\text{ + 210} \\ u_2\text{ = 0.75(547.5) + 210} \\ u_2\text{ = 410.625 + 210 } \\ u_2\text{ = 620.625} \end{gathered}[/tex]We will follow the above steps to find u3, u4 and u5
b) To find the value of the previous term:
New term = un, previous term = un-1
[tex]\begin{gathered} u_n=0.75u_{n-1}+\text{ 210 Make the previous term the subject of formula} \\ 0.75u_{n-1}\text{ = }u_n\text{ - 210 Divide through by 0.75} \\ u_{n-1}\text{ = }\frac{(u_n-210)}{0.75} \end{gathered}[/tex]The order of operation is to make the previous term the subject of the formula.
This done by bringing 210 to the other side of the equation: subtract 210 on both sides of the equation.
Then dividing both sides of the equation by the coefficient of the previous equation (0.75).
c) A recursive formula that generates the values from 13a in
reverse order has been derived in question (b)
[tex]u_{n-1}\text{ = }\frac{(u_n-210)}{0.75}[/tex]
