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.

Short 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-by-step solution

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

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

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.