To solve the given problem, let's start by examining the term [tex]\(a_k\)[/tex] provided:
[tex]\[ a_k = \frac{1}{k(k+1)} \][/tex]
First, it helps to simplify [tex]\(a_k\)[/tex]. We can use partial fraction decomposition to rewrite it:
[tex]\[
a_k = \frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}
\][/tex]
Solving for [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we get:
[tex]\[
1 = A(k+1) + Bk
\][/tex]
[tex]\[
1 = Ak + A + Bk
\][/tex]
[tex]\[
1 = (A + B)k + A
\][/tex]
From this equation, we find:
[tex]\[
A + B = 0 \quad \text{and} \quad A = 1
\][/tex]
Therefore, [tex]\(B = -1\)[/tex]. This leads us to:
[tex]\[
a_k = \frac{1}{k} - \frac{1}{k+1}
\][/tex]
Next, we need to find the sum of the series from [tex]\( k = 1 \)[/tex] to [tex]\( n \)[/tex]:
[tex]\[
\sum_{k=1}^n a_k = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)
\][/tex]
This is a telescoping series. When written out, most terms cancel each other:
[tex]\[
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right)
\][/tex]
All intermediate terms cancel, leaving us with:
[tex]\[
\sum_{k=1}^n a_k = 1 - \frac{1}{n+1}
\][/tex]
[tex]\[
\sum_{k=1}^n a_k = \frac{n}{n+1}
\][/tex]
We are now supposed to square the sum of the series:
[tex]\[
\left( \sum_{k=1}^n a_k \right)^2 = \left( \frac{n}{n+1} \right)^2
\][/tex]
[tex]\[
\left( \frac{n}{n+1} \right)^2 = \frac{n^2}{(n+1)^2}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{n^2}{(n+1)^2}}
\][/tex]
