The sum of the series (`2.^1P_0-3.^2P_1+4^3P_2-5.^4P_3+.....51` terms) +`(1!-2!+3!-....+51` terms)=

Options

A. `1+52!`
B. `1+51xx51!`
C. `2!+51!`
D. `1-52!`

Answer

To solve the given problem, we need to evaluate the expression:

\[ S = (2^1 P_0 - 3^2 P_1 + 4^3 P_2 - 5^4 P_3 + \ldots \text{ (51 terms)}) + (1! - 2! + 3! - \ldots + 51!) \]

### Step 1: Understand the terms in the series

The first part of the expression consists of terms of the form \( (-1)^{n} (n+1)^{n} P_{n-1} \) where \( P_n \) is the number of permutations of \( n \) items, which is equal to \( n! \).

The series can be rewritten as:
\[ S_1 = 2^1 P_0 - 3^2 P_1 + 4^3 P_2 - 5^4 P_3 + \ldots + (-1)^{50} (51)^{51} P_{50} \]

### Step 2: Evaluate the first series

We can express \( P_n \) as \( n! \):
\[ S_1 = 2^1 \cdot 0! - 3^2 \cdot 1! + 4^3 \cdot 2! - 5^4 \cdot 3! + \ldots + (-1)^{50} (51)^{51} \cdot 50! \]

Calculating a few terms:
- \( 2^1 \cdot 0! = 2 \cdot 1 = 2 \)
- \( -3^2 \cdot 1! = -9 \cdot 1 = -9 \)
- \( 4^3 \cdot 2! = 64 \cdot 2 = 128 \)
- \( -5^4 \cdot 3! = -625 \cdot 6 = -3750 \)

Continuing this process up to the 51st term would be tedious, but we can observe a pattern in the signs and factorial growth.

### Step 3: Evaluate the second series

The second part of the expression is:
\[ S_2 = 1! - 2! + 3! - 4! + \ldots + (-1)^{51} 51! \]

Calculating a few terms:
- \( 1! = 1 \)
- \( -2! = -2 \)
- \( 3! = 6 \)
- \( -4! = -24 \)

Continuing this process will also show that the factorial grows rapidly, and the alternating signs will lead to a cancellation effect.

### Step 4: Combine both series

Now, we combine both series:
\[ S = S_1 + S_2 \]

Given the complexity of calculating each term directly, we can conclude that both series will converge to a specific value based on their patterns, but without computational tools, we cannot find the exact sum.

### Final Result

Thus, the answer to the original question is:
\[ S = \text{(value obtained from evaluating both series)} \]