Find the coefficient of `x^5` in `(1+2x+3x^2...........)^(-3/2)`

Options

A. 21
B. 25
C. 26
D. None of these

Answer

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

### Step 1: Identify the series
The series \( 1 + 2x + 3x^2 + \ldots \) can be expressed as:
\[ S = \sum_{n=0}^{\infty} (n+1)x^n \]

### Step 2: Find the closed form of the series
This series can be recognized as the derivative of a geometric series. The sum of the geometric series \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \) for \( |x| < 1 \). Differentiating both sides gives:
\[ \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \]
Multiplying by \( x \):
\[ \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \]
Now, adding the first term (which is 1):
\[ S = 1 + \sum_{n=1}^{\infty} nx^n = 1 + \frac{x}{(1-x)^2} = \frac{1 - x + x}{(1-x)^2} = \frac{1}{(1-x)^2} \]

### Step 3: Rewrite the original expression
Now we can rewrite the original expression:
\[ (1 + 2x + 3x^2 + \ldots)^{-3/2} = \left(\frac{1}{(1-x)^2}\right)^{-3/2} = (1-x)^3 \]

### Step 4: Expand using the Binomial Theorem
Using the Binomial Theorem, we can expand \( (1-x)^3 \):
\[ (1-x)^3 = \sum_{k=0}^{3} \binom{3}{k} (-x)^k = 1 - 3x + 3x^2 - x^3 \]

### Step 5: Identify the coefficient of \( x^5 \)
From the expansion \( 1 - 3x + 3x^2 - x^3 \), we can see that the highest power of \( x \) is \( x^3 \). Therefore, there are no terms involving \( x^5 \).

### Conclusion
Thus, the coefficient of \( x^5 \) in the expansion is:
\[ \text{Coefficient of } x^5 = 0 \]