$^nC_m - 3^{n-1}C_m + 5^{n-2}C_m - 7^{n-3}C_m + \ldots + (2(n-m)-1)^nC_m$ is equal to:
Step-by-Step Solution
Key Concept: When the argument of a ratio equals $\pm\frac{\pi}{2}$, the locus is a circle with the two fixed points as diameter endpoints.
Given $\arg\left(\frac{z-2}{z+2}\right) = \pm\frac{\pi}{2}$, we rewrite as $\arg(z_1 - z_2) = \arg(z_1 - z_3) = \pm\frac{\pi}{2}$. This means the argument of the difference between $z$ and two fixed points $z_2, z_3$ differs by $\pm\frac{\pi}{2}$, indicating that $z$ lies on a circle with diameter connecting these points. The condition forces $z_1, z_2, z_3$ to form a right-angled triangle.
Correct Answer: I need to find the value of the series: $^nC_m - 3^{n-1}C_m + 5^{n-2}C_m - 7^{n-3}C_m + \ldots + (2(n-m)-1)^nC_m$
Let me analyze the pattern: