If `3+5x + 7x^2`…., `oo` term `s=44/9` , then x is equal to:

Options

A. `1//4`
B. `3//4`
C. `17//11`
D. None of these

Answer

To solve the problem step by step, we will derive the value of \( x \) from the given infinite series and its sum.

### Step 1: Write the series and its sum
The series is given as:
\[ S = 3 + 5x + 7x^2 + 9x^3 + \ldots \]
It is given that the sum \( S = \frac{44}{9} \).

### Step 2: Multiply the series by \( x \)
Multiply both sides of the equation by \( x \):
\[ xS = 3x + 5x^2 + 7x^3 + 9x^4 + \ldots \]

### Step 3: Subtract the two equations
Now, subtract \( xS \) from \( S \):
\[ S - xS = (3 + 5x + 7x^2 + 9x^3 + \ldots) - (3x + 5x^2 + 7x^3 + \ldots) \]
This simplifies to:
\[ S(1 - x) = 3 + (5x - 3x) + (7x^2 - 5x^2) + (9x^3 - 7x^3) + \ldots \]
\[ S(1 - x) = 3 + 2x + 2x^2 + 2x^3 + \ldots \]

### Step 4: Factor out the common terms
The right-hand side can be factored:
\[ S(1 - x) = 3 + 2(x + x^2 + x^3 + \ldots) \]
The series \( x + x^2 + x^3 + \ldots \) is a geometric series with first term \( x \) and common ratio \( x \):
\[ x + x^2 + x^3 + \ldots = \frac{x}{1 - x} \quad \text{(for } |x| < 1\text{)} \]
Thus, we have:
\[ S(1 - x) = 3 + 2 \cdot \frac{x}{1 - x} \]

### Step 5: Substitute \( S \) and solve for \( x \)
Substituting \( S = \frac{44}{9} \):
\[ \frac{44}{9}(1 - x) = 3 + \frac{2x}{1 - x} \]
Multiply through by \( 9(1 - x) \) to eliminate the fraction:
\[ 44(1 - x) = 27(1 - x) + 18x \]
Expanding both sides:
\[ 44 - 44x = 27 - 27x + 18x \]
Combine like terms:
\[ 44 - 44x = 27 - 9x \]
Rearranging gives:
\[ 44 - 27 = 44x - 9x \]
\[ 17 = 35x \]
Thus:
\[ x = \frac{17}{35} \]

### Step 6: Check the value of \( x \)
Since \( x \) must be less than 1 for the series to converge, \( \frac{17}{35} < 1 \) is valid.

### Conclusion
The value of \( x \) is:
\[ \boxed{\frac{17}{35}} \]