What is the coefficient of `x^(5)` in the expansion `( 1-2x + 3x^(2) - 4x^(3)+"………"oo)^(-5)` ?

Options

A. `( 10!)// ( 5!)^(2)`
B. `5^(-5)`
C. `5^(5)`
D. `10!//(6!)( 4!)`

Answer

To find the coefficient of \( x^5 \) in the expansion of \( (1 - 2x + 3x^2 - 4x^3 + \ldots)^{-5} \), we can follow these steps:

### Step 1: Identify the series
The series inside the parentheses can be recognized as a power series. It can be expressed as:
\[ 1 - 2x + 3x^2 - 4x^3 + \ldots = \frac{1}{(1 + 2x)^{2}} \]
This is derived from the formula for the sum of a geometric series.

### Step 2: Rewrite the expression
Thus, we can rewrite the original expression as:
\[ (1 - 2x + 3x^2 - 4x^3 + \ldots)^{-5} = \left( \frac{1}{(1 + 2x)^{2}} \right)^{-5} = (1 + 2x)^{10} \]

### Step 3: Use the Binomial Theorem
Now, we will use the Binomial Theorem to expand \( (1 + 2x)^{10} \). The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 1 \), \( b = 2x \), and \( n = 10 \):
\[ (1 + 2x)^{10} = \sum_{k=0}^{10} \binom{10}{k} (1)^{10-k} (2x)^k = \sum_{k=0}^{10} \binom{10}{k} 2^k x^k \]

### Step 4: Find the coefficient of \( x^5 \)
To find the coefficient of \( x^5 \), we need to look at the term where \( k = 5 \):
\[ \text{Coefficient of } x^5 = \binom{10}{5} 2^5 \]

### Step 5: Calculate the values
Now we calculate \( \binom{10}{5} \) and \( 2^5 \):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
\[ 2^5 = 32 \]

### Step 6: Multiply to find the final coefficient
Now, we multiply these two results:
\[ \text{Coefficient of } x^5 = 252 \times 32 = 8064 \]

### Final Answer
The coefficient of \( x^5 \) in the expansion of \( (1 - 2x + 3x^2 - 4x^3 + \ldots)^{-5} \) is \( 8064 \).

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