Questions: Suppose you make a deposit of P into a savings account that earns interest at a rate of 100 r % per year. a. Show that if interest is compounded once per year, then the balance after t years is B(t)=P(1+r)^(1/2). b. If interest is compounded m times per year, then the balance after t years is B(T)=P(1+r/m)^(mt) In the limit m -> infinity, the compounding is said to be continuous. Show that with continuous compounding, the balance after t years is B(t)=Pe. a. Begin by stating the compound interest formula. B(t)=P(1+r/n)^(nt), where P is the principal, r is the annual rate of interest as a decimal, t is the number of years the amount is deposited for, and n is the number of times the interest is compounded per year.

Suppose you make a deposit of P into a savings account that earns interest at a rate of 100 r % per year.
a. Show that if interest is compounded once per year, then the balance after t years is B(t)=P(1+r)^(1/2).
b. If interest is compounded m times per year, then the balance after t years is B(T)=P(1+r/m)^(mt) In the limit m -> infinity, the compounding is said to be continuous. Show that with continuous compounding, the balance after t years is B(t)=Pe.
a. Begin by stating the compound interest formula.
B(t)=P(1+r/n)^(nt), where P is the principal, r is the annual rate of interest as a decimal, t is the number of years the amount is deposited for, and n is the number of times the interest is compounded per year.

Transcript text: Suppose you make a deposit of \$P into a savings account that earns interest at a rate of $100 \mathrm{r} \%$ per year. a. Show that if interest is compounded once per year, then the balance after $t$ years is $B(t)=P(1+r)^{\frac{1}{2}}$. b. If interest is compounded $m$ times per year, then the balance after $t$ years is $B(T)=P\left(1+\frac{\mathrm{r}}{\mathrm{m}}\right)^{\mathrm{mt}}$ In the limit $\mathrm{m} \rightarrow \infty$, the compounding is said to be continuous. Show that with continuous compounding, the balance after t years is $\mathrm{B}(t)=\mathrm{Pe}$. a. Begin by stating the compound interest formula. $\mathrm{B}(\mathrm{t})=\mathrm{P}\left(1+\frac{r}{n}\right)^{\mathrm{nt}}$, where P is the principal, r is the annual rate of interest as a decimal, t is the number of years the amount is deposited for, and $n$ is the number of times the interest is compounded per year.

Solution

Solution Steps

Step 1: Annual Compounding Formula

Using the compound interest formula for annual compounding, we have: \[ B(t) = P(1 + r)^t \] Substituting $ P = 1000 $, $ r = 0.05 $, and $ t = 10 $: \[ B(10) = 1000(1 + 0.05)^{10} \]

Step 2: Calculate Annual Balance

Calculating the balance after 10 years with annual compounding: \[ B(10) = 1000(1.05)^{10} \approx 1628.89 \]

Step 3: Compounding $ m $ Times Per Year

Using the compound interest formula for $ m $ times per year: \[ B(t) = P\left(1 + \frac{r}{m}\right)^{mt} \] Substituting $ P = 1000 $, $ r = 0.05 $, $ m = 12 $, and $ t = 10 $: \[ B(10) = 1000\left(1 + \frac{0.05}{12}\right)^{12 \cdot 10} \]

Step 4: Calculate Balance for $ m $ Compounding

Calculating the balance after 10 years with monthly compounding: \[ B(10) = 1000\left(1 + \frac{0.05}{12}\right)^{120} \approx 1647.01 \]

Step 5: Continuous Compounding Formula

For continuous compounding, the formula is: \[ B(t) = Pe^{rt} \] Substituting $ P = 1000 $, $ r = 0.05 $, and $ t = 10 $: \[ B(10) = 1000e^{0.05 \cdot 10} \]

Step 6: Calculate Continuous Balance

Calculating the balance after 10 years with continuous compounding: \[ B(10) = 1000e^{0.5} \approx 1648.72 \]

Final Answer

For part a: $ r = 0.05, n = 1 $
For part b: $ B(t) = Pe^{rt} $ with $ B(t) = Pe $ when $ t = 1 $.
Thus, the final answer is:
$\boxed{B(t) = Pe}$ for continuous compounding.

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