A grandmother offers her granddaughter Jennifer two choices for a saving plan: O

Question

A grandmother offers her granddaughter Jennifer two choices for a saving plan: Option 1: On 30 December of year 1, the grandmother opens a bank account with 200$ for Jennifer. On 29 December of each year, she pays 'interest', that is, she deposits an amount that corresponds to 2.9% of the account balance on that day into the account. Immediately after the 'interest payment', she deposits an additional 170$ to the account. Let A(n) denote the amount of money in the account on 31 December of year n. Option 2: On 30 December of year 1, the grandmother opens a bank account with 120$ for Jennifer. Starting in year 2, she deposits 190$ into this account on 2 January of every year. On 29 December of each year, she pays 'interest', that is, she deposits an amount that corresponds to 5% of the account balance on that day to the account. Let B(n) denote the amount of money in the account on 31 December of year n. (a) Find a recursive definition for A_n. (You may find it helpful to draw a timeline.) (b) Find a recursive definition for B_n. (You may find it helpful to draw a timeline.)

Explanation / Answer

Solution

Part (a)

Amount as on 301201(i.e., 30th December of year 1) = 200

Amount as on 291202(i.e., 29th December of year 2) = (200 x 1.029) + 170

Amount as on 291203

= {(200 x 1.029) + 170}x1.029 +170 = = {(200 x 1.0292) + (170 x1.029) +170}

Similarly, Amount as on 291204 = {(200 x 1.0293) + (170 x 1.0292) + (170 x 1.029) + 170}

Or, in general, Amount as on 3112 of nth year

= An = {(200 x 1.029n - 1) + (170 x 1.029n - 2) + (170 x 1.029n - 3) + …+ (170 x 1.029) + 170}

= (200 x 1.029n - 1) + 170(1.029n – 1 – 1)/(1.029 – 1) [Applying the rule for summing a Geometric Progression of (n - 1) terms with first term as 170 and common ratio = 1.029]

= (200 x 1.029n - 1) + 170(1.029n – 1 – 1)/(0.029)

= (1/0.029) {(200 x 1.029n – 1 x 0.029) + 170(1.029n – 1 – 1)}

= [(1.029n – 1 x 175.8) – 170]/0.029 ANSWER

Part (b)

Amount as on 301201(i.e., 30th December of year 1) = 120

Amount as on 020102(i.e., 2 January of year 2) = (120 + 190)

Amount as on 291202 = {(120 + 190) x 1.05}

Amount as on 020103 = {(120 + 190) x 1.05} + 190

Amount as on 291203 = [{(120 + 190) x 1.05} + 190] x 1.05

= [(120 x 1.052) + (190 x 1.052) + (190 x 1.05)]

Similarly,

Amount as on 291204 = [(120 x 1.053) + (190 x 1.053) + (190 x 1.052) + (190 x 1.05)]

Or, in general, Amount as on 3112 of nth year

= Bn

= [(120 x 1.05n - 1) + (190 x 1.05n - 1) + (190 x 1.05n - 2) + …… + (190 x 1.05)]

= [(120 x 1.05n - 1) + {190 x 1.05(1.05n – 1 – 1)/(1.05 - 1)] [Applying the rule for summing a

   Geometric Progression of (n - 1) terms with first term as (190 x 1.05) and common ratio = 1.05]

= [(120 x 1.05n - 1) + {190 x 21(1.05n – 1 – 1)]

= [(120 x 1.05n - 1) + (3990 x 1.05n – 1) – 3990]

= [(4110 x 1.05n - 1) – 3990] ANSWER

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