`P = 1 - (3)/(1!) + 9/(2!) - (27)/(3!) + .......`
`Q = 1+ (4)/(1!) +(16)/(2!) +(64)/(3!) + ........`
`R=log_(e)3+((log_e3)^2)/(2!)+((log_e3)^3)/(3!)+.....`
The ascending order of P,Q,R

Options

A. Q,P,R
B. P,R,Q
C. P,Q,R
D. R,P,Q

Answer

To solve the problem, we need to evaluate the series \( P \), \( Q \), and \( R \) and then compare their values to determine the ascending order.

### Step 1: Evaluate \( P \)
The series for \( P \) is given by:
\[ P = 1 - \frac{3}{1!} + \frac{9}{2!} - \frac{27}{3!} + \ldots \]
This series can be recognized as the expansion of \( e^x \) where \( x = -3 \):
\[ P = e^{-3} \]

### Step 2: Evaluate \( Q \)
The series for \( Q \) is given by:
\[ Q = 1 + \frac{4}{1!} + \frac{16}{2!} + \frac{64}{3!} + \ldots \]
This series can be recognized as the expansion of \( e^x \) where \( x = 4 \):
\[ Q = e^{4} \]

### Step 3: Evaluate \( R \)
The series for \( R \) is given by:
\[ R = \log_e 3 + \frac{(\log_e 3)^2}{2!} + \frac{(\log_e 3)^3}{3!} + \ldots \]
This series can be recognized as the expansion of \( e^x \) where \( x = \log_e 3 \):
\[ R = e^{\log_e 3} = 3 \]

### Step 4: Compare the values of \( P \), \( Q \), and \( R \)
Now we have:
- \( P = e^{-3} \)
- \( Q = e^{4} \)
- \( R = 3 \)

To compare these values:
- The value of \( e^{-3} \) is approximately \( \frac{1}{e^3} \approx \frac{1}{20.0855} \approx 0.0498 \), which is less than 1.
- The value of \( e^{4} \) is approximately \( 54.5981 \), which is significantly greater than 3.
- The value of \( R \) is exactly 3.

### Conclusion: Ascending Order
From our evaluations:
- \( P \approx 0.0498 < R = 3 < Q \approx 54.5981 \)

Thus, the ascending order of \( P \), \( Q \), and \( R \) is:
\[ P < R < Q \]