Binomial Theorem
Binomial Theorem
star_batch_jee_advanced_2025
Grade None
Question:
$\sum_{r=1}^{n} \frac{(-1)^{r-1}{}^nC_r 0 - y^f}{r} =$
$x + \frac{x^2}{2} + \frac{x^3}{3} + .... + \frac{x^n}{n}$
$\frac{1-x}{1} + \frac{1-x^2}{2} + \frac{1-x^3}{3} + .... + \frac{1-x^n}{n}$
$(x-1) + \frac{x^2-1}{2} + \frac{x^3-1}{3} + .... + \frac{x^n-1}{n}$
$n - \frac{x}{1} - \frac{x^2}{2} - \frac{x^3}{3} - .... - \frac{x^n}{n}$
Step-by-Step Solution
Key Concept: The locus of points equidistant from two endpoints forms a circle with diameter equal to the distance between those endpoints.
The midpoint of the segment $OB$ lies on the center of a circle. Since $O$ is at the origin and $B$ corresponds to $z_0(1+1)$, the midpoint is at $\frac{z_0(1+1)}{2}$ and the radius equals $\frac{|z_0|}{2}$. Therefore, the required equation of the circle is $\left|z - \frac{z_0(1+1)}{2}\right| = \frac{|z_0|}{2}$.
Correct Answer: I need to find the sum $\sum_{r=1}^{n} \frac{(-1)^{r-1}{}^nC_r x^r}{r}$ (interpreting the garbled notation in the question).
This is a standard binomial series problem. Let me work through it systematically.
Starting with the binomial expansion