The sum to (n+1) terms of the series
`(C_(0))/(2)-(C_(1))/(3)+(C_(2))/(4)-(C_(3))/(5)+…` is

Options

A. `(1)/(n+1)`
B. `(1)/(n+2)`
C. `(1)/(n(n+1))`
D. none of these

Answer

To find the sum to (n + 1) terms of the series

\[ S = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \ldots \]

we can follow these steps:

### Step 1: Understand the series
The series consists of binomial coefficients \(C_k\) divided by consecutive integers starting from 2. The general term can be expressed as \((-1)^k \frac{C_k}{k + 2}\).

### Step 2: Use the Binomial Theorem
Recall the binomial expansion of \((1 - x)^n\):

\[ (1 - x)^n = \sum_{k=0}^{n} C_k (-x)^k \]

### Step 3: Multiply by \(x\) and Integrate
To find the desired sum, we multiply both sides of the binomial expansion by \(x\) and integrate from 0 to 1:

\[ \int_0^1 x(1 - x)^n \, dx = \int_0^1 \left( C_0 x - C_1 x^2 + C_2 x^3 - C_3 x^4 + \ldots \right) \, dx \]

### Step 4: Evaluate the left-hand side
The left-hand side can be evaluated using integration by parts or a known formula for the integral:

\[ \int_0^1 x(1 - x)^n \, dx = \frac{1}{(n + 1)(n + 2)} \]

### Step 5: Evaluate the right-hand side
The right-hand side becomes:

\[ \int_0^1 \left( C_0 x - C_1 x^2 + C_2 x^3 - C_3 x^4 + \ldots \right) \, dx \]

This results in:

\[ \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \ldots \]

### Step 6: Set the two sides equal
Now we have:

\[ \frac{1}{(n + 1)(n + 2)} = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \ldots \]

### Step 7: Final Result
Thus, we conclude:

\[ \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \ldots = \frac{1}{(n + 1)(n + 2)} \]