A freshly passed out B.Tech. Student from PDEU starts her job at a software company with a salary of Rs. 50,000 and is promised that at the end of each year her salary will be double her salary of the previous year, with an extra increment of Rs. 10,000 for each year she has been with company.

Construct and solve a recurrence relation to find her salary for her nth year of employment.

Let \( S_n \) denote the salary of the student in her \( n \)-th year of employment. We need to construct a recurrence relation for \( S_n \) based on the given conditions.

### Problem Breakdown:
- Initial salary (\( S_1 \)) in the first year: Rs. 50,000.
- Each year, her salary doubles from the previous year, with an additional increment of Rs. 10,000 for each year she has been with the company.

#### Given Information:
1. Initial condition: \( S_1 = 50,000 \).
2. For each \( n \geq 2 \), the salary is doubled compared to the previous year, plus an additional Rs. 10,000 multiplied by the number of years she has been with the company.

Thus, the recurrence relation can be written as:

\[
S_n = 2S_{n-1} + 10,000 \times (n-1) \quad \text{for} \quad n \geq 2.
\]

### Base Case:
\[
S_1 = 50,000.
\]

### Recurrence Relation:
\[
S_n = 2S_{n-1} + 10,000 \times (n-1) \quad \text{for} \quad n \geq 2.
\]

### Solution:

To find \( S_n \), let’s solve this recurrence relation step by step:

- For \( n = 2 \):
  \[
  S_2 = 2S_1 + 10,000 \times 1 = 2 \times 50,000 + 10,000 = 100,000 + 10,000 = 110,000.
  \]

- For \( n = 3 \):
  \[
  S_3 = 2S_2 + 10,000 \times 2 = 2 \times 110,000 + 20,000 = 220,000 + 20,000 = 240,000.
  \]

- For \( n = 4 \):
  \[
  S_4 = 2S_3 + 10,000 \times 3 = 2 \times 240,000 + 30,000 = 480,000 + 30,000 = 510,000.
  \]

- For \( n = 5 \):
  \[
  S_5 = 2S_4 + 10,000 \times 4 = 2 \times 510,000 + 40,000 = 1,020,000 + 40,000 = 1,060,000.
  \]

So, the salary in the 5th year is Rs. 1,060,000. You can compute the salary for any \( n \)-th year using this recurrence relation.

Would you like a detailed derivation of the closed-form solution, or any further explanation?

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**Relative questions:**
1. What if the yearly increment were Rs. 15,000 instead of Rs. 10,000? How would the recurrence change?
2. What if the salary only increased by Rs. 10,000 each year without doubling?
3. How would the solution differ if the initial salary was Rs. 60,000 instead of Rs. 50,000?
4. How can this recurrence relation be solved using characteristic equations?
5. How does this recurrence relation compare to compound interest growth?

**Tip:** Solving recurrence relations often gives insight into future values in sequences, allowing predictions beyond just a few terms.

A freshly passed out B.Tech. student from PDEU starts her job at a software company with a salary of Rs. 50,000 and is promised that at the end of each year her salary will be double her salary of the previous year, with an extra increment of Rs. 10,000 for each year she has been with the company. Construct and solve a recurrence relation to find her salary for her nth year of employment.

Solve the recurrence relation for a student's salary at a software company, where the initial salary is Rs. 50,000, doubling every year with an additional increment of Rs. 10,000 per year of employment. Learn step-by-step how to find the salary for the nth year, starting from the base case and applying the recurrence formula.

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