Find the value of `.^20 C_0-(.^20C_1)/2+(.^20C_2)/3-(.^20C_3)/4+...`

Explanation

To find the value of the series

\[ S = \binom{20}{0} - \frac{\binom{20}{1}}{2} + \frac{\binom{20}{2}}{3} - \frac{\binom{20}{3}}{4} + \ldots \]

we can express the general term of the series as:

\[ T_R = \frac{\binom{20}{R}}{R+1} (-1)^R \]

where \( R \) ranges from 0 to 20. Thus, we can write:

\[ S = \sum_{R=0}^{20} T_R = \sum_{R=0}^{20} \frac{\binom{20}{R}}{R+1} (-1)^R \]

### Step 1: Rewrite the General Term
Using the identity

\[ \frac{\binom{N}{R}}{R+1} = \frac{1}{N+1} \binom{N+1}{R+1} \]

we can rewrite our series:

\[ S = \sum_{R=0}^{20} \frac{\binom{20}{R}}{R+1} (-1)^R = \frac{1}{21} \sum_{R=0}^{20} \binom{21}{R+1} (-1)^R \]

### Step 2: Change the Index of Summation
Now, we can change the index of summation by letting \( k = R + 1 \). Then when \( R = 0 \), \( k = 1 \) and when \( R = 20 \), \( k = 21 \):

\[ S = \frac{1}{21} \sum_{k=1}^{21} \binom{21}{k} (-1)^{k-1} \]

### Step 3: Factor out the Negative Sign
We can factor out the negative sign from the summation:

\[ S = -\frac{1}{21} \sum_{k=0}^{21} \binom{21}{k} (-1)^k \]

### Step 4: Evaluate the Summation
The summation \( \sum_{k=0}^{21} \binom{21}{k} (-1)^k \) is equal to \( (1 - 1)^{21} = 0 \).

### Step 5: Substitute Back into the Equation
Substituting this back into our expression for \( S \):

\[ S = -\frac{1}{21} \cdot 0 = 0 \]

### Step 6: Adjust for the Missing Term
Since we factored out \( k=0 \) from the summation, we need to add back \( \binom{21}{0} \):

\[ S = -\frac{1}{21} (0 - 1) = \frac{1}{21} \]

### Final Result
Thus, the value of the series is

\[ \boxed{\frac{1}{21}} \]