If `3/(4!)+4/(5!)+5/(6!)+.....+50 term=1/(3!)-1/((k+3)!)`, then sum of coefficients in the expansion `(1+2x_1+3x_2+.....+100x_(100))^k` is:

Options

A. (a) `(5050)^(49)`
B. (b) `(5050)^(51)`
C. (c) `(5050)^(52)`
D. (d) `(5050)^(50)`

Answer

To solve the problem step by step, we will analyze the given series and the expression we need to evaluate.

### Step 1: Identify the Series
The series given is:
\[ S = \frac{3}{4!} + \frac{4}{5!} + \frac{5}{6!} + \ldots \]
We need to find the 50th term of this series.

### Step 2: Find the General Term
The general term of the series can be expressed as:
\[ \text{Term}_n = \frac{n + 2}{(n + 3)!} \]
where \( n \) starts from 1. For \( n = 1 \), we have the first term as \( \frac{3}{4!} \).

### Step 3: Find the 50th Term
To find the 50th term, we substitute \( n = 50 \):
\[ \text{Term}_{50} = \frac{50 + 2}{(50 + 3)!} = \frac{52}{53!} \]

### Step 4: Set Up the Equation
According to the problem, we have:
\[ S = \frac{1}{3!} - \frac{1}{(k + 3)!} \]
Now, we need to express \( S \) in terms of factorials to compare with the right-hand side.

### Step 5: Simplify the Series
We can rewrite the terms in the series:
\[ S = \sum_{n=1}^{50} \frac{n + 2}{(n + 3)!} = \sum_{n=1}^{50} \left( \frac{n + 3}{(n + 3)!} - \frac{1}{(n + 3)!} \right) \]
This gives:
\[ S = \sum_{n=1}^{50} \frac{1}{(n + 2)!} - \sum_{n=1}^{50} \frac{1}{(n + 3)!} \]

### Step 6: Evaluate the Sums
The first sum can be evaluated as:
\[ \sum_{n=1}^{50} \frac{1}{(n + 2)!} = \frac{1}{3!} + \frac{1}{4!} + \ldots + \frac{1}{52!} \]
The second sum is:
\[ \sum_{n=1}^{50} \frac{1}{(n + 3)!} = \frac{1}{4!} + \frac{1}{5!} + \ldots + \frac{1}{53!} \]

### Step 7: Combine the Sums
Now we can combine the two sums:
\[ S = \left( \frac{1}{3!} + \frac{1}{4!} + \ldots + \frac{1}{52!} \right) - \left( \frac{1}{4!} + \frac{1}{5!} + \ldots + \frac{1}{53!} \right) \]
This simplifies to:
\[ S = \frac{1}{3!} - \frac{1}{53!} \]

### Step 8: Solve for \( k \)
From the equation:
\[ \frac{1}{3!} - \frac{1}{(k + 3)!} = \frac{1}{3!} - \frac{1}{53!} \]
We can equate:
\[ \frac{1}{(k + 3)!} = \frac{1}{53!} \]
Thus, \( k + 3 = 53 \) which gives:
\[ k = 50 \]

### Step 9: Find the Sum of Coefficients
Now we need to find the sum of coefficients in the expansion of:
\[ (1 + 2x_1 + 3x_2 + \ldots + 100x_{100})^k \]
Substituting \( k = 50 \), we need to evaluate:
\[ (1 + 2 + 3 + \ldots + 100)^{50} \]

### Step 10: Calculate the Sum of First 100 Natural Numbers
The sum of the first 100 natural numbers is given by:
\[ \text{Sum} = \frac{n(n + 1)}{2} = \frac{100 \times 101}{2} = 5050 \]

### Final Step: Raise to the Power of \( k \)
Thus, the sum of coefficients is:
\[ 5050^{50} \]

### Final Answer
The sum of coefficients in the expansion is:
\[ \boxed{5050^{50}} \]