`1+1/4 + (1xx3)/(4xx8) + (1xx3xx5)/(4xx8xx12) + "….."` is equal to

Options

A. a. `sqrt(2)`
B. b. `(1)/(sqrt(2))`
C. c. `sqrt(3)`
D. d. `(1)/(sqrt(3))`

Explanation

To solve the series \( S = 1 + \frac{1}{4} + \frac{1 \times 3}{4 \times 8} + \frac{1 \times 3 \times 5}{4 \times 8 \times 12} + \ldots \) step by step, we can relate it to the binomial theorem.

### Step 1: Identify the series
We can express the series as:
\[ S = 1 + \frac{1}{4} + \frac{1 \cdot 3}{4 \cdot 8} + \frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12} + \ldots \]
This series can be recognized as a form of the binomial expansion.

### Step 2: Relate to the binomial theorem
The binomial theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k \]
where \( \binom{n}{k} = \frac{n(n-1)(n-2)\ldots(n-k+1)}{k!} \).

### Step 3: Identify \( n \) and \( x \)
We can compare our series with the binomial expansion. The first term is 1, which matches. The second term is \( \frac{1}{4} \), which corresponds to \( nx \). Thus, we have:
\[ nx = \frac{1}{4} \]

### Step 4: Find the next term
The next term in our series is \( \frac{1 \cdot 3}{4 \cdot 8} \). We can express this as:
\[ \frac{n(n-1)}{2} x^2 = \frac{3}{32} \]
From the previous step, we have \( nx = \frac{1}{4} \). Squaring this gives:
\[ n^2 x^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]

### Step 5: Substitute and simplify
Now substituting \( x^2 \) into the equation for the next term:
\[ \frac{n(n-1)}{2} \cdot \frac{1}{16} = \frac{3}{32} \]
Multiplying both sides by 32:
\[ 16n(n-1) = 3 \]
This simplifies to:
\[ n(n-1) = \frac{3}{16} \]

### Step 6: Solve for \( n \)
We can rearrange this to form a quadratic equation:
\[ n^2 - n - \frac{3}{16} = 0 \]
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + \frac{12}{16}}}{2} = \frac{1 \pm \sqrt{\frac{28}{16}}}{2} = \frac{1 \pm \frac{\sqrt{7}}{2}}{2} \]
This gives:
\[ n = \frac{1 + \sqrt{7}/2}{2} \quad \text{or} \quad n = \frac{1 - \sqrt{7}/2}{2} \]

### Step 7: Find \( x \)
From \( nx = \frac{1}{4} \), we can find \( x \):
\[ x = \frac{1}{4n} \]

### Step 8: Substitute back into the series
Now substituting \( n \) and \( x \) back into the binomial expansion gives us the value of the series:
\[ S = (1 + x)^n \]

### Step 9: Calculate \( S \)
Substituting the values of \( n \) and \( x \) into the expression gives us:
\[ S = \left(1 - \frac{1}{2}\right)^{-1/2} = 2^{1/2} = \sqrt{2} \]

### Final Result
Thus, the value of the series is:
\[ \sqrt{2} \]