If `a_(K)=(1)/(K(K+1))` for `K=1, 2, 3....n`, then `(sum_(K=1)^(n)a_(K))^(2)=`

Options

A. `(n)/(n+1)`
B. `(n^(2))/((n+1)^(2))`
C. `(n^(4))/((n+1)^(4))`
D. `(n^(6))/((n+1)^(6))`

Answer

To solve the problem, we need to find the expression for \((\sum_{K=1}^{n} a_{K})^{2}\) where \(a_{K} = \frac{1}{K(K+1)}\).

### Step-by-step Solution:

1. **Identify the expression for \(a_{K}\)**:
\[ a_{K} = \frac{1}{K(K+1)} \]

2. **Decompose \(a_{K}\) into partial fractions**:
We can express \(a_{K}\) as:
\[ a_{K} = \frac{1}{K} - \frac{1}{K+1} \]

3. **Set up the summation**:
We need to compute:
\[ \sum_{K=1}^{n} a_{K} = \sum_{K=1}^{n} \left( \frac{1}{K} - \frac{1}{K+1} \right) \]

4. **Evaluate the summation**:
This is a telescoping series. When we expand it, we get:
\[ \sum_{K=1}^{n} \left( \frac{1}{K} - \frac{1}{K+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \ldots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \]
Most terms cancel out, leaving us with:
\[ 1 - \frac{1}{n+1} = \frac{n}{n+1} \]

5. **Square the result**:
Now we square the result of the summation:
\[ \left( \sum_{K=1}^{n} a_{K} \right)^{2} = \left( \frac{n}{n+1} \right)^{2} = \frac{n^{2}}{(n+1)^{2}} \]

### Final Result:
Thus, the final answer is:
\[ \left( \sum_{K=1}^{n} a_{K} \right)^{2} = \frac{n^{2}}{(n+1)^{2}} \]