Compound interest

The formula to calculate compound interest is:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

Where:

• A = the amount of money accumulated after interest, including principal
• P = the principal amount (the initial money invested)
• r = annual interest rate (decimal)
• n = number of times the interest is compounded per year
• t = the time the money is invested or borrowed for, in years

Ibrahimu invested 20,000/= at 6% compound interest for 5 years. How much does he have at the end of 5 years?

Given:

• P = 20,000
• r = 6% = 0.06
• n = 1 (since interest is compounded annually)
• t = 5 years

Using the compound interest formula:

$A = 20,000 \times \left(1 + \frac{0.06}{1}\right)^{1 \times 5} = 20,000 \times (1.06)^5 = 20,000 \times 1.338225$

$A = 26,764.50$

Answer: At the end of 5 years, Ibrahimu will have 26,764.50/=.

At the beginning of each year, Martha invests 10,000/= at 5% compound interest. How much does she have at the end of the 10th year?

Each year, Martha makes a new investment, and each investment earns interest for a different number of years. Let's break down her investments:

The 1st investment has had 10 years of interest, so it's:

$10,000 \times (1.05)^{10} = 10,000 \times 1.628894 = 16,288.94$

The 2nd investment has had 9 years of interest, so it's:

$10,000 \times (1.05)^9 = 10,000 \times 1.551328 = 15,513.28$

The 3rd investment has had 8 years of interest, so it's:

$10,000 \times (1.05)^8 = 10,000 \times 1.477455 = 14,774.55$

Following this pattern, the 10th investment has had only 1 year of interest, so it's:

$10,000 \times (1.05)^1 = 10,000 \times 1.05 = 10,500$

Now, we add all these amounts together to get the total after 10 years:

$Total = 10,000 \times (1.05)^{10} + 10,000 \times (1.05)^9 + ... + 10,000 \times (1.05)^1$

$Total = 16,288.94 + 15,513.28 + 14,774.55 + ... + 10,500$

This is a geometric series with first term $10,000 \times 1.05$ and common ratio $1.05$. The sum of the series is given by:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Where:

• a = $10,000 \times 1.05$ (the first term)
• r = 1.05 (the common ratio)
• n = 10 (the number of terms)

Using the formula for the sum of a geometric series:

$S_{10} = 10,500 \times \frac{1 - 1.05^{10}}{1 - 1.05} = 10,500 \times \frac{1 - 1.628894}{-0.05}$

$S_{10} = 10,500 \times \frac{-0.628894}{-0.05} = 10,500 \times 12.57788$

$S_{10} \approx 132,060.74$

Answer: Martha will have approximately 132,060.74/= at the end of the 10th year.