$^nC_1(1 + \frac{1}{2}){}^nC_2 + (1 + \frac{1}{2} + \frac{1}{3}){}^nC_3 - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}){}^nC_4 + .... + (-1)^{n-1}(1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n}){}^nC_n =$
Step-by-Step Solution
Key Concept: A complex number is purely imaginary if and only if its sum with its conjugate equals zero, which translates to a real part condition.
Given $\angle OAP = \frac{\pi}{2}$, the quantity $\frac{z - z_0}{z_0}$ is purely imaginary. This means $\frac{z - z_0}{z_0} + \overline{\left(\frac{z - z_0}{z_0}\right)} = 0$. Simplifying: $\frac{z}{z_0} + \frac{\bar{z}}{\bar{z_0}} = 2$, which gives $2\operatorname{Re}\left(\frac{z}{z_0}\right) = 2$, hence $\operatorname{Re}\left(\frac{z}{z_0}\right) = 1$.
Correct Answer: I need to find the value of:
$$\sum_{k=1}^{n} (-1)^{k-1}\left(\sum_{j=1}^{k}\frac{1}{j}\right)\binom{n}{k}$$
Let me denote $H_k = 1 + \frac{1}{2