If `(1 - k) (1 + 2x + 4x^(2) + 8x^(3) + 16x^(4) + 32x^(5)) = 1 - k^(6)`, where `k ne 1` then the value of `(k)/(x)` is

Options

A. 2
B. 4
C. `1//2`
D. `1//4`

Answer

To solve the equation \((1 - k)(1 + 2x + 4x^2 + 8x^3 + 16x^4 + 32x^5) = 1 - k^6\), we can follow these steps:

### Step 1: Recognize the Series
The series \(1 + 2x + 4x^2 + 8x^3 + 16x^4 + 32x^5\) is a geometric progression (GP) where:
- The first term \(a = 1\)
- The common ratio \(r = 2x\)
- The number of terms \(n = 6\)

### Step 2: Use the Formula for the Sum of a GP
The sum \(S_n\) of the first \(n\) terms of a GP can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Substituting the values:
\[ S_6 = \frac{1(1 - (2x)^6)}{1 - 2x} = \frac{1 - 64x^6}{1 - 2x} \]

### Step 3: Substitute the Sum Back into the Equation
Now we substitute this sum back into the original equation:
\[ (1 - k) \cdot \frac{1 - 64x^6}{1 - 2x} = 1 - k^6 \]

### Step 4: Cross Multiply
Cross-multiplying gives:
\[ (1 - k)(1 - 64x^6) = (1 - k^6)(1 - 2x) \]

### Step 5: Expand Both Sides
Expanding both sides:
\[ 1 - 64x^6 - k + 64kx^6 = 1 - k^6 - 2x + 2kx^6 \]

### Step 6: Rearranging Terms
Rearranging the equation leads to:
\[ -64x^6 + 64kx^6 + 2x - 1 + k - k^6 = 0 \]

### Step 7: Equate Coefficients
For the equation to hold for all \(x\), the coefficients of \(x^6\) must match. This gives us:
\[ 64k = 2 \quad \text{and} \quad k - k^6 = 0 \]

### Step 8: Solve for \(k\)
From \(64k = 2\):
\[ k = \frac{2}{64} = \frac{1}{32} \]

### Step 9: Substitute \(k\) into the Second Equation
Substituting \(k\) back into \(k - k^6 = 0\):
\[ \frac{1}{32} - \left(\frac{1}{32}\right)^6 = 0 \]
This is satisfied since both terms are equal.

### Step 10: Find \(\frac{k}{x}\)
Since \(k = 2x\) from the earlier steps, we can substitute:
\[ \frac{k}{x} = \frac{2x}{x} = 2 \]

Thus, the final answer is:
\[ \frac{k}{x} = 2 \]

---