The problem describes an investment of Rs. 1,00,000, where at the end of each year, 20% interest is added to the account balance, and no money is withdrawn.

Step 1: Define the recurrence relation.

Let $A_n$ represent the amount in the account at the end of year $n$.

• Initial deposit: $A_0 = 1,00,000$ (this is the amount at the beginning).
• Every year, 20% of the amount gets added to the account. Hence, the amount in the account at the end of year $n$ is 120% (or 1.20 times) of the amount at the end of the previous year $n-1$.

So, the recurrence relation is: $A_n = 1.20 \cdot A_{n-1}, \quad \text{for} \, n \geq 1$ with the initial condition: $A_0 = 1,00,000$

Step 2: Solve the recurrence relation.

This is a geometric sequence, where each term is multiplied by 1.20 to get the next term.

The general solution to this recurrence relation is: $A_n = A_0 \cdot (1.20)^n$

Substitute $A_0 = 1,00,000$: $A_n = 1,00,000 \cdot (1.20)^n$

Final Expression:

The amount in the account after $n$ years is: $A_n = 1,00,000 \cdot (1.20)^n$

This equation gives the amount in the account at the end of year $n$, considering 20% interest compounded annually.

Would you like to explore this further or ask for details on the solution?

Here are some relative questions you might find useful:

1. How does the recurrence relation change with different interest rates?
2. What would happen if there were additional deposits each year?
3. How would withdrawals affect the recurrence relation and solution?
4. How do we calculate the interest compounded quarterly or monthly?
5. What is the difference between simple interest and compound interest in this context?

Tip: Always verify the compounding period when solving interest problems, as it can affect the recurrence relation significantly.