IF `1+(1/2)+(1/3)+…….+(1/20)=k` then what is the value of `(1/4)+(1/6)+(1/8)+…..+(1/40)`?

Options

A. `k/2`
B. 2k
C. `((k-1))/2`
D. `((k+1))/2`

Answer

To solve the problem step by step, we need to find the value of the series \( \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} \) given that \( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{20} = k \).

### Step 1: Rewrite the series
We can express the series \( \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} \) in a more manageable form. Notice that all the denominators in this series are even numbers. We can factor out \( \frac{1}{2} \) from each term:

\[ \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} = \frac{1}{2} \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{20} \right) \]

### Step 2: Identify the inner series
The inner series \( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{20} \) can be rewritten by separating the first term:

\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{20} = \left( 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{20} \right) - 1 = k - 1 \]

### Step 3: Substitute back into the equation
Now we can substitute this back into our expression from Step 1:

\[ \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} = \frac{1}{2} \left( k - 1 \right) \]

### Step 4: Final expression
Thus, we can express the final answer as:

\[ \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} = \frac{k - 1}{2} \]

### Conclusion
The value of the series \( \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \ldots + \frac{1}{40} \) is \( \frac{k - 1}{2} \).

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