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

Options

A. A) `1//4`
B. B) `1//16`
C. C) 1
D. D) `1//28`

Answer

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

### Step 1: Identify the series
The expression \( 1 + 2x + 3x^2 + 4x^3 + \ldots \) can be recognized as a power series. Specifically, it can be rewritten in a more manageable form. The series can be expressed as:
\[ \sum_{n=0}^{\infty} (n+1)x^n \]
This series can be derived from the formula for the sum of a geometric series.

### Step 2: Recognize the series as a function
The series \( \sum_{n=0}^{\infty} (n+1)x^n \) can be derived from the function:
\[ \frac{1}{(1-x)^2} \]
This is because the derivative of \( \frac{1}{1-x} \) gives \( \frac{1}{(1-x)^2} \), and multiplying by \( x \) gives us the series we have.

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

### Step 4: Expand the function using the binomial theorem
Now we need to expand \( \frac{1}{\sqrt{1-x^2}} \) using the binomial series expansion:
\[ (1-x^2)^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-x^2)^n \]
This simplifies to:
\[ \sum_{n=0}^{\infty} \binom{n+\frac{1}{2}}{n} x^{2n} \]

### Step 5: Find the coefficient of \( x^4 \)
We want the coefficient of \( x^4 \) in this expansion. The term \( x^4 \) corresponds to \( n=2 \) (since \( 2n = 4 \)):
\[ \text{Coefficient of } x^4 = \binom{2+\frac{1}{2}}{2} = \binom{\frac{5}{2}}{2} \]

### Step 6: Calculate the binomial coefficient
Calculating \( \binom{\frac{5}{2}}{2} \):
\[ \binom{\frac{5}{2}}{2} = \frac{\frac{5}{2} \cdot \frac{3}{2}}{2!} = \frac{\frac{15}{4}}{2} = \frac{15}{8} \]

### Conclusion
Thus, the coefficient of \( x^4 \) in the expansion of \( (1 + 2x + 3x^2 + 4x^3 + \ldots)^{\frac{1}{2}} \) is \( \frac{15}{8} \).