If `sum_(r=1)^(oo) (C^r)/(r!) =1` then C =

Options

A. 1
B. 2
C. e
D. `log_e2`

Answer

To solve the equation \( \sum_{r=1}^{\infty} \frac{C^r}{r!} = 1 \), we can follow these steps:

### Step 1: Write out the series
The series can be expressed as:
\[ \frac{C^1}{1!} + \frac{C^2}{2!} + \frac{C^3}{3!} + \ldots \]

### Step 2: Add 1 to both sides
To manipulate the series, we can add 1 to both sides:
\[ 1 + \left( \frac{C^1}{1!} + \frac{C^2}{2!} + \frac{C^3}{3!} + \ldots \right) = 1 + 1 \]
This simplifies to:
\[ 1 + \sum_{r=1}^{\infty} \frac{C^r}{r!} = 2 \]

### Step 3: Recognize the series as an exponential function
The left-hand side can be recognized as the Taylor series expansion for \( e^C \):
\[ e^C = 1 + \sum_{r=1}^{\infty} \frac{C^r}{r!} \]
Thus, we have:
\[ e^C = 2 \]

### Step 4: Take the natural logarithm of both sides
To solve for \( C \), we take the natural logarithm:
\[ \ln(e^C) = \ln(2) \]
Using the property of logarithms, this simplifies to:
\[ C = \ln(2) \]

### Conclusion
Thus, the value of \( C \) is:
\[ C = \ln(2) \]

---