JEE Mathematics 2023 Question with Solution

Given: The series is

$$1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 10$$

Find: The value of $k$.

From the solution, the coefficients after the first term follow the pattern

$$4, 8, 13, 19, \ldots$$

whose successive differences are

$$4, 5, 6, \ldots$$

So for powers starting from $\frac{1}{k}$, the general coefficient can be written accordingly, and the solution concludes after simplification that

$$k = 2$$

Therefore, the value of $k$ is $2$.

Given:

$$1 + \frac{4}{k} + \frac{8}{k^2} + \frac{13}{k^3} + \frac{19}{k^4} + \ldots = 10$$

Find: $k$

The provided solution states these intermediate forms:

1. Subtracting $1$ gives

$$9 = 4k^2 + 8k^3 + 13k^4 + 19k^5 + \dots$$

2. Multiplying by $k$ gives

$$9k = 4k^3 + 8k^4 + 13k^5 + \dots$$

3. It then rewrites the series as

$$S = 4k + 4k^2 + 5k^3 + 6k^4 + \dots$$

4. Finally, after simplification, it concludes

$$k = 2$$

The provided working appears abbreviated and has notation inconsistencies, but the final conclusion on the solution's is that the required value is $2$.