`1+(2)/(4) + (2.5)/(4.8)+(2.5.8)/(4.8.12)+(2.5.8.11)/(4.8.12.16)+…..=`

Options

A. `4^(-2//3)`
B. `root(5)(16)`
C. `root(3)(4)`
D. `4^(2//3)`

Answer

To solve the series

\[ 1 + \frac{2}{4} + \frac{2 \cdot 5}{4 \cdot 8} + \frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12} + \frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16} + \ldots \]

we can analyze the pattern in the terms and relate it to the binomial theorem.

### Step 1: Identify the pattern in the series

The series can be rewritten as:

\[ 1 + \frac{2}{4} + \frac{2 \cdot 5}{4 \cdot 8} + \frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12} + \frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16} + \ldots \]

Notice that the numerators consist of products of terms that increase by 3, starting from 2, while the denominators are products of terms that increase by 4, starting from 4.

### Step 2: Relate the series to the binomial theorem

The binomial theorem states that:

\[ (1 - x)^{-n} = \sum_{k=0}^{\infty} \frac{n(n+1)(n+2)\ldots(n+k-1)}{k!} x^k \]

We can relate our series to this expansion.

### Step 3: Compare terms

The first term of the series is 1, which corresponds to \( (1 - x)^{-n} \) when \( k = 0 \).

The second term \( \frac{2}{4} \) can be matched with \( nx \), where \( n \) is some constant we need to determine.

### Step 4: Set up equations

From the second term:

\[ nx = \frac{2}{4} \implies nx = \frac{1}{2} \]

From the third term:

\[ \frac{n(n+1)}{2} x^2 = \frac{2 \cdot 5}{4 \cdot 8} \]

Calculating the right-hand side:

\[ \frac{2 \cdot 5}{4 \cdot 8} = \frac{10}{32} = \frac{5}{16} \]

### Step 5: Solve for \( n \) and \( x \)

Now we have two equations:

1. \( nx = \frac{1}{2} \)
2. \( \frac{n(n+1)}{2} x^2 = \frac{5}{16} \)

From the first equation, we can express \( n \) in terms of \( x \):

\[ n = \frac{1}{2x} \]

Substituting this into the second equation:

\[ \frac{\left(\frac{1}{2x}\right)\left(\frac{1}{2x} + 1\right)}{2} x^2 = \frac{5}{16} \]

### Step 6: Simplify and solve

This simplifies to:

\[ \frac{1}{4x^2} \left(\frac{1 + 2x}{2}\right) x^2 = \frac{5}{16} \]

This leads to:

\[ \frac{1 + 2x}{8} = \frac{5}{16} \]

Cross-multiplying gives:

\[ 2(1 + 2x) = 5 \implies 2 + 4x = 5 \implies 4x = 3 \implies x = \frac{3}{4} \]

### Step 7: Find \( n \)

Substituting \( x = \frac{3}{4} \) back into \( n = \frac{1}{2x} \):

\[ n = \frac{1}{2 \cdot \frac{3}{4}} = \frac{2}{3} \]

### Step 8: Final Result

Now we have \( n = \frac{2}{3} \) and \( x = \frac{3}{4} \). The series can be expressed as:

\[ (1 - x)^{-n} = (1 - \frac{3}{4})^{-\frac{2}{3}} = (1/4)^{-\frac{2}{3}} = 4^{\frac{2}{3}} \]

Thus, the value of the series is:

\[ \boxed{4^{\frac{2}{3}}} \]