Lewis saves KSh 200 at the beginning of each month for 12 months. If the bank offers a simple interest rate of 1/4% per annum, calculate the total interest earned by December.

The principal amount saved each month, $P = \text{KSh } 200$.
The annual simple interest rate, $R = \frac{1}{4}\% = 0.25\%$.
The period of saving is from January 2018 to December 2018, which is 12 months.

The formula for simple interest is $I = \frac{P \times R \times T}{100}$, where $T$ is the time in years.

Lewis saves KSh 200 at the beginning of each month.

• The January deposit earns interest for 12 months ($T = \frac{12}{12}$ years).
• The February deposit earns interest for 11 months ($T = \frac{11}{12}$ years).
• The March deposit earns interest for 10 months ($T = \frac{10}{12}$ years).
• ...
• The December deposit earns interest for 1 month ($T = \frac{1}{12}$ years).

The total interest is the sum of the interest earned by each monthly deposit.

$$I_{total} = \frac{200 \times 0.25 \times \frac{12}{12}}{100} + \frac{200 \times 0.25 \times \frac{11}{12}}{100} + \dots + \frac{200 \times 0.25 \times \frac{1}{12}}{100}$$

Factor out the common terms:

$$I_{total} = \frac{200 \times 0.25}{100} \times \left( \frac{12}{12} + \frac{11}{12} + \dots + \frac{1}{12} \right)$$ $$I_{total} = \frac{50}{100} \times \frac{1}{12} \times (12 + 11 + \dots + 1)$$ $$I_{total} = 0.5 \times \frac{1}{12} \times (12 + 11 + \dots + 1)$$

The sum of the first 12 natural numbers is given by the formula $\frac{n(n+1)}{2}$.
For $n=12$:

$$12 + 11 + \dots + 1 = \frac{12(12+1)}{2} = \frac{12 \times 13}{2} = 6 \times 13 = 78$$

Substitute this sum back into the total interest equation:

$$I_{total} = 0.5 \times \frac{1}{12} \times 78$$ $$I_{total} = \frac{0.5 \times 78}{12}$$ $$I_{total} = \frac{39}{12}$$ $$I_{total} = \frac{13}{4}$$ $$I_{total} = 3.25$$

The total interest received during the 12 months is KSh 3.25.

$$\text{KSh } 3.25$$