The problem is about constructing and solving a recurrence relation for the salary of a B.Tech graduate, who starts with an initial salary of ₹50,000 and has a specific salary growth rule.

Problem Breakdown:

• Initial salary (Year 1): ₹50,000.
• Salary increment rule:
  • At the end of each year, the salary doubles compared to the previous year.
  • Additionally, there is a fixed increment of ₹10,000 for each year worked.

Recurrence Relation Setup:

Let $S_n$ represent her salary at the end of the $n$-th year.

• Base case: $S_1 = 50,000$
• Recurrence relation: According to the problem, at the end of each year:
  • Salary doubles: $2S_{n-1}$
  • Plus a fixed increment of ₹10,000 for each year worked.

Thus, the recurrence relation becomes: $S_n = 2S_{n-1} + 10,000$ Where $S_1 = 50,000$.

Solving the Recurrence Relation:

We now need to solve this recurrence relation to find a general formula for $S_n$.

Let me solve this step-by-step for the salary in the $n$-th year.The solution to the recurrence relation is:

$S_n = 30,000 \cdot 2^n - 10,000$

This formula represents the salary at the end of the $n$-th year.

• For $n = 1$, it simplifies to the initial salary: $S_1 = 30,000 \cdot 2^1 - 10,000 = 50,000$
• For larger $n$, you can use this formula to calculate the salary after any given year.

Do you want further details or clarifications? Here are some related questions to explore:

1. What will the salary be at the end of the 5th year?
2. How does the salary growth compare if the fixed increment was ₹15,000 instead of ₹10,000?
3. How would the salary change if the initial salary was ₹40,000 instead of ₹50,000?
4. Can the recurrence relation be solved using other methods like iteration?
5. How would the recurrence relation look if the salary increased by 150% instead of doubling?

Tip: You can check how rapidly the salary grows by substituting different values of $n$ into the formula.