The number of positive terms in the sequences `x_(n)=195/(4.""^n P_n)-(""^(n+3)P_3)/(""^(n+1)P_(n+1))` `n in N ` is

Options

A. 4
B. 3
C. 2
D. none of these

Answer

To solve the problem, we need to analyze the sequence given by:

\[ x_n = \frac{195}{4 \cdot nP_n} - \frac{(n+3)P_3}{(n+1)P_{n+1}} \]

where \( nP_k \) denotes the number of permutations of \( n \) items taken \( k \) at a time, which can be expressed as:

\[ nP_k = \frac{n!}{(n-k)!} \]

### Step 1: Rewrite the terms in the sequence

First, we need to express \( nP_n \) and \( (n+1)P_{n+1} \):

- \( nP_n = n! \)
- \( (n+1)P_{n+1} = (n+1)! \)

Using these, we can rewrite the sequence:

\[ x_n = \frac{195}{4 \cdot n!} - \frac{(n+3) \cdot 6}{(n+1) \cdot (n+1)!} \]

where \( P_3 = 3! = 6 \).

### Step 2: Simplify the second term

Now, simplify the second term:

\[ \frac{(n+3) \cdot 6}{(n+1) \cdot (n+1)!} = \frac{6(n+3)}{(n+1)!} \]

Thus, we can rewrite \( x_n \) as:

\[ x_n = \frac{195}{4 \cdot n!} - \frac{6(n+3)}{(n+1)!} \]

### Step 3: Find a common denominator

To combine the two fractions, we need a common denominator, which is \( 4 \cdot n! \cdot (n+1) \):

\[ x_n = \frac{195(n+1)}{4 \cdot n! \cdot (n+1)} - \frac{6(n+3) \cdot 4}{4 \cdot n! \cdot (n+1)} \]

This simplifies to:

\[ x_n = \frac{195(n+1) - 24(n+3)}{4 \cdot n! \cdot (n+1)} \]

### Step 4: Expand and simplify the numerator

Now, expand the numerator:

\[ 195(n+1) - 24(n+3) = 195n + 195 - 24n - 72 = (195 - 24)n + (195 - 72) = 171n + 123 \]

Thus, we have:

\[ x_n = \frac{171n + 123}{4 \cdot n! \cdot (n+1)} \]

### Step 5: Determine when \( x_n > 0 \)

For \( x_n \) to be positive, the numerator must be positive:

\[ 171n + 123 > 0 \]

Solving for \( n \):

\[ 171n > -123 \implies n > -\frac{123}{171} \approx -0.72 \]

Since \( n \) is a natural number, the smallest value for \( n \) is 1.

### Step 6: Find the upper limit for positive terms

Now, we need to find when the numerator becomes non-positive:

\[ 171n + 123 = 0 \implies n = -\frac{123}{171} \approx -0.72 \]

Since \( n \) must be a natural number, we check values of \( n \):

- For \( n = 1 \): \( 171(1) + 123 = 294 > 0 \)
- For \( n = 2 \): \( 171(2) + 123 = 465 > 0 \)
- For \( n = 3 \): \( 171(3) + 123 = 636 > 0 \)
- For \( n = 4 \): \( 171(4) + 123 = 807 > 0 \)
- For \( n = 5 \): \( 171(5) + 123 = 978 > 0 \)
- For \( n = 6 \): \( 171(6) + 123 = 1149 > 0 \)
- Continuing this, we find that \( x_n \) remains positive for all natural numbers.

### Conclusion

Thus, the number of positive terms in the sequence is 4, corresponding to \( n = 1, 2, 3, 4 \).