If `C_(0), C_(1), C_(2),..., C_(n)` are binomial coefficients
in the expansion of `(1 + x)^(n), ` then the value of
`C_(0) + (C_(1))/(2) + (C_(2))/(3) + (C_(3))/(4) +...+ (C_(n))/(n+1)` is

Options

A. `(2^(n+1) - 1)/(n+1)`
B. `(2^(n) - 1)/(n)`
C. `(2^(n-1) - 1)/(n - 1)`
D. `(2^(n+1) - 1)/(n + 2)`

Answer

To solve the problem, we need to find the value of

\[ C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \frac{C_3}{4} + \ldots + \frac{C_n}{n+1} \]

where \(C_k\) are the binomial coefficients from the expansion of \((1 + x)^n\).

### Step 1: Write the binomial expansion
The binomial expansion of \((1 + x)^n\) is given by:

\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]

where \(C_k = \binom{n}{k}\).

### Step 2: Integrate the expansion
To find the required sum, we can integrate the expansion from 0 to 1:

\[ \int_0^1 (1 + x)^n \, dx \]

### Step 3: Calculate the integral
The integral can be computed as follows:

\[ \int (1 + x)^n \, dx = \frac{(1 + x)^{n+1}}{n + 1} + C \]

Evaluating this from 0 to 1 gives:

\[ \left[ \frac{(1 + x)^{n+1}}{n + 1} \right]_0^1 = \frac{(1 + 1)^{n+1}}{n + 1} - \frac{(1 + 0)^{n+1}}{n + 1} \]

This simplifies to:

\[ \frac{2^{n+1}}{n + 1} - \frac{1}{n + 1} = \frac{2^{n+1} - 1}{n + 1} \]

### Step 4: Relate the integral to the sum
Now, we can express the integral in terms of the binomial coefficients:

\[ \int_0^1 (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) \, dx = C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} \]

### Step 5: Set the equations equal
Thus, we have:

\[ C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \ldots + \frac{C_n}{n + 1} = \frac{2^{n+1} - 1}{n + 1} \]

### Final Result
Therefore, the value of

\[ C_0 + \frac{C_1}{2} + \frac{C_2}{3} + \frac{C_3}{4} + \ldots + \frac{C_n}{n+1} = \frac{2^{n+1} - 1}{n + 1} \]