The problem provided defines two functions $f(n)$ and $g(n)$, and you're asked to evaluate $g(100)$. Here's how we break it down:

Function Definitions:

1. $f(n) = \sum_{i=1}^{2n-1} (-1)^{i-1} \cdot \frac{i}{\binom{2n}{i}}$, where $\binom{2n}{i}$ is the binomial coefficient.
2. $g(n) = f(n) + \frac{1}{n+1}$

Step-by-step Solution:

We are tasked with finding $g(100)$, which involves calculating $f(100)$ first and then using the formula for $g(n)$.

Step 1: Calculate $f(100)$

To compute $f(100)$, we need to evaluate the sum from $i = 1$ to $199$ for: $f(100) = \sum_{i=1}^{199} (-1)^{i-1} \cdot \frac{i}{\binom{200}{i}}$

This is a fairly complex sum involving binomial coefficients and alternating signs, but can be computed step by step.

Step 2: Calculate $g(100)$

Once $f(100)$ is known, the value of $g(100)$ is: $g(100) = f(100) + \frac{1}{101}$

Let me calculate the sum for $f(100)$ and then plug it into the formula for $g(100)$. I'll work on that calculation now.The value of $g(100)$ is $1.0$.

Let me know if you need further explanation or have any other questions!

Here are 5 related questions to expand on this topic:

1. What properties of binomial coefficients are useful in simplifying sums like this?
2. How does alternating signs in a series affect its convergence?
3. What is the significance of the binomial coefficient in combinatorics?
4. Can this type of series be evaluated using integration techniques?
5. How would the formula change if the function $f(n)$ had a different summation range?

Tip: When dealing with binomial coefficients and alternating series, look for patterns or symmetries that can help simplify computations.