Q. 89 - Calculus

Question

 Suppose you invest $100.00 in a bank that pays you 5% interest compounded annually. The balance in the account after k years is given by $a_k=100{(1-0.05)}^k$To the nearest cent, determine the first five terms of the sequence, starting at k = 0. What does k = 0 mean in practical terms ?Determine whether the sequence is bounded. Determine whether the sequence is increasing, decreasing, or not monotonic.

Step-by-Step Solution

Answer

As the value of k increases, the balance a, increases. Hence, the sequence is not bounded. Also, the value of a, increases as k increases, hence, the sequence is an increasing sequence.

Step 1. Given

The Initial Investment in a bank is $100.

The balance after k years is given by a $a_k=100{(1-0.05)}^k$.

Step 2. Calculation

$$\begin{gathered} \mathrm{To} \mathrm{find} \mathrm{the} \mathrm{first} \mathrm{five} \mathrm{terms} \mathrm{of} \mathrm{the} \mathrm{sequence}, \mathrm{substitute} k = 0,1,2,3,4 in a = 100{(1+0.05)}^k. \mathrm{Substituting} k=0, \\ {a}_0 = 100{(1+0.05)}^0 =100{(1+0.05)}^0 \mathrm{Hence}, a_0 = 100 \\ \mathrm{Substituting} k=1, \\ {a}_1 = 100{(1+0.05)}^1 \\ =100{(1+0.05)}^1 \\ \mathrm{Hence}, a_1 = 105 \\ \mathrm{Substituting} k=2, \\ {a}_2 = 100{(1+0.05)}^2 \\ =100{(1+0.05)}^2 \\ \mathrm{Hence}, a_2 = 110.25 \\ \mathrm{Substituting} k=3, \\ {a}_3 = 100{(1+0.05)}^3 \\ =100{(1+0.05)}^3 \\ \mathrm{Hence}, a_3 = 115.76 \\ \mathrm{Substituting} k=4, \\ {a}_4 = 100{(1+0.05)}^4 \\ =100{(1+0.05)}^4 \\ \mathrm{Hence}, a_4 = 121.55 \end{gathered}$$

Step 3. Result

As the value of k increases, the balance a, increases. Hence, the sequence is not bounded. Also, the value of a, increases as k increases, hence, the sequence is an increasing sequence.