`1/((20-a)(40-a))+1/((40-a)(60-a))+. . .+1/((180-a)(200-a))=1/256` then find the value of `a`

Options

A. 8
B. 6
C. 4
D. 5

Answer

To solve the equation

\[ \frac{1}{(20-a)(40-a)} + \frac{1}{(40-a)(60-a)} + \cdots + \frac{1}{(180-a)(200-a)} = \frac{1}{256} \]

we will first analyze the left-hand side.

### Step 1: Identify the series
The left-hand side consists of a series of fractions. The general term can be expressed as:

\[ \frac{1}{(n-a)(n+20-a)} \]

where \( n \) takes values from 20 to 180 in steps of 20 (i.e., 20, 40, 60, ..., 180).

### Step 2: Rewrite the series
We can rewrite the series:

\[ \sum_{n=20, n \text{ step } 20}^{180} \frac{1}{(n-a)(n+20-a)} \]

This can be simplified using partial fractions:

\[ \frac{1}{(n-a)(n+20-a)} = \frac{1}{20} \left( \frac{1}{n-a} - \frac{1}{n+20-a} \right) \]

### Step 3: Apply the partial fractions
Substituting this back into the sum gives:

\[ \frac{1}{20} \left( \left( \frac{1}{20-a} - \frac{1}{40-a} \right) + \left( \frac{1}{40-a} - \frac{1}{60-a} \right) + \cdots + \left( \frac{1}{180-a} - \frac{1}{200-a} \right) \right) \]

### Step 4: Simplify the series
Notice that this is a telescoping series, where most terms cancel out:

\[ \frac{1}{20} \left( \frac{1}{20-a} - \frac{1}{200-a} \right) \]

### Step 5: Set the equation
Now, we set this equal to \(\frac{1}{256}\):

\[ \frac{1}{20} \left( \frac{1}{20-a} - \frac{1}{200-a} \right) = \frac{1}{256} \]

### Step 6: Solve for \( a \)
Multiplying both sides by 20 gives:

\[ \frac{1}{20-a} - \frac{1}{200-a} = \frac{20}{256} = \frac{5}{64} \]

Now, finding a common denominator on the left side:

\[ \frac{(200-a) - (20-a)}{(20-a)(200-a)} = \frac{5}{64} \]

This simplifies to:

\[ \frac{180}{(20-a)(200-a)} = \frac{5}{64} \]

### Step 7: Cross-multiply
Cross-multiplying gives:

\[ 180 \cdot 64 = 5(20-a)(200-a) \]

Calculating \( 180 \cdot 64 \):

\[ 11520 = 5(20-a)(200-a) \]

### Step 8: Expand and rearrange
Expanding the right side:

\[ 11520 = 5(4000 - 220a + a^2) \]

Dividing through by 5:

\[ 2304 = 4000 - 220a + a^2 \]

Rearranging gives:

\[ a^2 - 220a + (4000 - 2304) = 0 \]

This simplifies to:

\[ a^2 - 220a + 1696 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ a = \frac{220 \pm \sqrt{220^2 - 4 \cdot 1 \cdot 1696}}{2 \cdot 1} \]

Calculating the discriminant:

\[ 220^2 = 48400, \quad 4 \cdot 1696 = 6784 \]

Thus:

\[ \sqrt{48400 - 6784} = \sqrt{41616} = 204 \]

So:

\[ a = \frac{220 \pm 204}{2} \]

Calculating the two possible values:

1. \( a = \frac{424}{2} = 212 \)
2. \( a = \frac{16}{2} = 8 \)

### Step 10: Determine valid solutions
Since \( a \) must be less than 20 (to keep the denominators positive), we discard \( a = 212 \).

Thus, the valid solution is:

\[ \boxed{8} \]