Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
Let $x_1, x_2, x_3, ..., x_{2018}$ be real numbers different from 1, such that $x_1 + x_2 + ... + x_{2018} = 1$ and $\frac{x_1}{1-x_1} + \frac{x_2}{1-x_2} + ... + \frac{x_{2018}}{1-x_{2018}} = 1$ then the value of $\frac{x_1^2}{1-x_1} + \frac{x_2^2}{1-x_2} + ... + \frac{x_{2018}^2}{1-x_{2018}}$ is equal to ____.
Step-by-Step Solution
Key Concept: Decomposing each term as $-(x_r+1) + \frac{1}{1-x_r}$ allows telescoping and simplification of the sum.
The sum $\sum_{r=1}^{2018} \frac{x_r^2}{1-x_r}$ is rewritten by expressing $\frac{x_r^2}{1-x_r} = -(x_r+1) + \frac{1}{1-x_r}$. Summing from $r=1$ to $2018$ and using telescoping properties, the first part gives $-1$ and the second part relates to the harmonic sum. The calculation yields that the sum equals $-1+1=0$ when properly simplified.
Correct Answer: 0