The sum of the series $^nC_1 + \frac{1+2}{2}{}^nC_2 + \frac{1+2+3}{3}{}^nC_3 + .... + \frac{1+2+3+...+n}{n}{}^nC_n$ is equal to:
$\frac{1}{2}(n^{2n-1}C_n + 2^nC_n)$
$\frac{1}{2}(n^{2n-1}C_n + 2^nC_n - 1)$
$\frac{1}{((n+1)^{2n-1}C_n - 1)}$
$\frac{1}{2}(2^{2n}C_n + 2^nC_n - 2)$
Step-by-Step Solution
Key Concept: Converting the logarithmic equation to exponential form and using component-wise matching of real and imaginary parts determines the relationship between $x$, $\theta$, and $k$.
Step 1: Define the initial equation and the variable $k$
The problem begins by setting up an equation that defines the variable $k$ in two equivalent forms involving complex logarithms and inverse trigonometric functions.
$$i \log\left(\frac{x-1}{x+1}\right) = -\pi + 2\tan^{-1} x = k$$
This establishes two fundamental relations:
1. $i \log\left(\frac{x-1}{x+1}\right) = k$
2. $-\pi + 2\tan^{-1} x = k$
Step 2: Convert to exponential form and introduce $\theta$
From the first relation, $i \log\left(\frac{x-1}{x+1}\right) = k$, we can write $\log\left(\frac{x-1}{x+1}\right) = \frac{k}{i} = -ik$.
This implies that $\frac{x-1}{x+1} = e^{-ik}$.
Taking the reciprocal of both sides, we obtain $\frac{x+1}{x-1} = e^{ik}$.
The solution then introduces an auxiliary variable $\theta$ such that this expression is equated to $e^{i\theta}$:
$$\frac{x+1}{x-1} = e^{i\theta}$$
where $\theta$ is explicitly defined as $\theta = k - \pi + 2\tan^{-1} x$.
Step 3: Solve a related complex equation
The solution proceeds by presenting and solving an auxiliary complex equation involving $x$ and $\theta$.
It states: "Expanding $x + i(x\sin\theta - \cos\theta) = x\cos\theta + i\sin\theta$ and equating real and imaginary parts yields $x = \cot\frac{\theta}{2}$ and $\theta = 2\cot^{-1} x$."
The identities derived from this auxiliary equation are used in subsequent steps.
Step 4: Determine the values of $k$
The final step involves using the established relationships and known trigonometric identities to find the possible values of $k$.
The solution presents the equation:
$$k + \pi = 2[\cot^{-1} x + \tan^{-1} x] - 2\left(\frac{\pi}{2}\right)$$
Using the fundamental inverse trigonometric identity $\cot^{-1} x + \tan^{-1} x = \frac{\pi}{2}$, we substitute this into the equation:
$$k + \pi = 2\left(\frac{\pi}{2}\right) - 2\left(\frac{\pi}{2}\right)$$
$$k + \pi = \pi - \pi$$
$$k + \pi = 0$$
Based on the provided solution, this ultimately leads to the conclusion regarding $k$.
The solution states that this calculation gives $k = \pi$ or $k = 0$.
The final answer is $k = \pi$ or $k = 0$.
Correct Answer: I need to find the sum of the series:
$$\sum_{r=1}^{n} \frac{1+2+...+r}{r} \binom{n}{r}$$
First, let me simplify the general term. The sum $1+2+...+r = \frac{r(r+1)}{2