To solve this problem, let's break it down step by step.
### Step 1: Understand the given series
The given series is:
[tex]\[ a_k = \frac{1}{k(k+1)} \][/tex]
We need to sum this series from [tex]\( k = 1 \)[/tex] to [tex]\( k = n \)[/tex]:
[tex]\[ \sum_{k=1}^n a_k = \sum_{k=1}^n \frac{1}{k(k+1)} \][/tex]
### Step 2: Simplify the expression for [tex]\( a_k \)[/tex]
Notice that [tex]\( a_k \)[/tex] can be simplified into partial fractions:
[tex]\[ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \][/tex]
### Step 3: Sum the series
The series [tex]\( \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) \)[/tex] is a telescoping series. Therefore, most terms will cancel out:
[tex]\[ \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \][/tex]
After cancellation, we are left with:
[tex]\[ 1 - \frac{1}{n+1} = \frac{n}{n+1} \][/tex]
So the sum of the series is:
[tex]\[ \sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1} \][/tex]
### Step 4: Square this sum
Now, we need to square the sum:
[tex]\[ \left( \sum_{k=1}^n a_k \right)^2 = \left( \frac{n}{n+1} \right)^2 = \frac{n^2}{(n+1)^2} \][/tex]
### Step 5: Compare with given options
Now let's compare this with the given options:
1. [tex]\(\frac{n}{n+1}\)[/tex]
2. [tex]\(\frac{n^2}{(n+1)^2}\)[/tex]
3. [tex]\(\frac{n^4}{(n+1)^4}\)[/tex]
4. [tex]\(\frac{n^6}{(n+1)^6}\)[/tex]
The correct option is:
[tex]\[ \frac{n^2}{(n+1)^2} \][/tex]
Thus, the answer is:
[tex]\[ \boxed{\frac{n^2}{(n+1)^2}} \][/tex]
