If, |x|<1,Find the coefficient of` x^n` in the expansion of `(1+ 2x + 3x^2 + 4x^3 + …..)^(1/2)`

Answer

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

### Step 1: Recognize the series
The series \( 1 + 2x + 3x^2 + 4x^3 + \ldots \) can be expressed as:
\[ \sum_{k=0}^{\infty} (k+1)x^k \]
This is a power series where the coefficient of \( x^k \) is \( k+1 \).

### Step 2: Identify the closed form of the series
The series can be simplified using the formula for the sum of a geometric series. We know that:
\[ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \quad \text{for } |x| < 1 \]
Differentiating this series with respect to \( x \) gives:
\[ \sum_{k=1}^{\infty} kx^{k-1} = \frac{1}{(1-x)^2} \]
Multiplying both sides by \( x \):
\[ \sum_{k=1}^{\infty} kx^k = \frac{x}{(1-x)^2} \]
Now, adding \( \sum_{k=0}^{\infty} x^k \) to this series:
\[ \sum_{k=0}^{\infty} (k+1)x^k = \sum_{k=0}^{\infty} x^k + \sum_{k=1}^{\infty} kx^k = \frac{1}{1-x} + \frac{x}{(1-x)^2} \]
Combining these gives:
\[ \sum_{k=0}^{\infty} (k+1)x^k = \frac{1}{1-x} + \frac{x}{(1-x)^2} = \frac{1}{(1-x)^2} \]

### Step 3: Substitute into the original expression
Now, we can rewrite the original expression:
\[ (1 + 2x + 3x^2 + 4x^3 + \ldots)^{1/2} = \left(\frac{1}{(1-x)^2}\right)^{1/2} = \frac{1}{1-x} \]

### Step 4: Expand the expression
The expression \( \frac{1}{1-x} \) can be expanded using the binomial series:
\[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \quad \text{for } |x| < 1 \]

### Step 5: Identify the coefficient of \( x^n \)
From the expansion \( \sum_{n=0}^{\infty} x^n \), we see that the coefficient of \( x^n \) is \( 1 \).

### Final Answer
Thus, the coefficient of \( x^n \) in the expansion of \( (1 + 2x + 3x^2 + 4x^3 + \ldots)^{1/2} \) is:
\[ \boxed{1} \]