smc-1007-60-Recursion

Tina inherits $60 000 and invests it in an account earning interest at a rate of 0.5% per month. Each month, immediately after the interest has been paid, Tina withdraws $800.

The amount in the account immediately after the `n`th withdrawal can be determined using the recurrence relation

`A_n = A_(n - 1)(1.005) - 800`,

where `n = 1, 2, 3, …` and `A_0 = 60\ 000`

Use the recurrence relation to find the amount of money in the account immediately after the third withdrawal. (2 marks)

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Calculate the amount of interest earned in the first three months. (2 marks)

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Calculate the amount of money in the account immediately after the 94th withdrawal. (3 marks)

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Show Answers Only

`$58\ 492.49`
`$892.49`
`$187.85`

Show Worked Solution

a.	`A_1`	`= 60\ 000(1.005) – 800 = $59\ 500`
	`A_2`	`= 59\ 500(1.005) – 800 = $58\ 997.50`
	`A_3`	`= 58\ 997.50(1.005) – 800 = $58\ 492.49`

b. `text{Amount (not interest)}`

`= 60\ 000 – (3 xx 800)`

`= $57\ 600`

`:.\ text(Interest earned in 3 months)`

`= A_3 – 57\ 600`

`= 58\ 492.49 – 57\ 600`

`= $892.49`

c. `A_1 = 60\ 000(1.005) – 800`

`A_2`	`= [60\ 000(1.005) – 800](1.005) – 800`
	`= 60\ 000(1.005)^n – 800(1.005 + 1)`
`vdots`
`A_n`	`= 60\ 000(1.005)^n – 800(1 + 1.005 + … + 1.005^(n – 1))`

`A_94`	`= 60\ 000(1.005)^94 – 800\ underbrace((1 + 1.005 + … + 1.005^93))_(text(GP where)\ a = 1,\ r = 1.005,\ n = 94)`
	`= 60\ 000(1.005)^94 – 800 ((1(1.005^94 – 1))/(1.005 – 1))`
	`= 60\ 000(1.005)^94 – 160\ 000(1.005^94 – 1)`
	`= $187.85`