Let's simplify the expression step by step. The expression seems to be a sum of two parts:

1. \( S = \sum_{k=1}^{N} \frac{1}{k(k+1)} \)
2. The second part appears as \( (N+1) \sum_{k=1}^{N} \frac{1}{k(k+1)} \).

So we can write the complete expression as:

\[
S + (N+1) S
\]

First, we will simplify \( S \):

\[
S = \sum_{k=1}^{N} \frac{1}{k(k+1)}
\]

We can use partial fraction decomposition:

\[
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
\]

Thus, the sum becomes:

\[
S = \sum_{k=1}^{N} \left( \frac{1}{k} - \frac{1}{k+1} \right)
\]

This sum is telescoping:

\[
S = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{N} - \frac{1}{N+1} \right)
\]

Most terms will cancel out, leaving:

\[
S = 1 - \frac{1}{N+1} = \frac{N}{N+1}
\]

Now substituting back into the full expression:

\[
S + (N+1) S = S + (N+1) \cdot \frac{N}{N+1}
\]

Calculating this gives us:

\[
S + (N+1) \cdot \frac{N}{N+1} = \frac{N}{N+1} + N = \frac{N + N(N+1)}{N+1} = \frac{N + N^2 + N}{N + 1} = \frac{N^2 + 2N}{N + 1}
\]

Finally, we arrive at the simplified result:

\[
\frac{N^2 + 2N}{N + 1}
\]

This can also be rewritten:

\[
\frac{N(N + 2)}{N + 1}
\]

Thus, the simplified form of your expression is:

\[
\frac{N(N + 2)}{N + 1}
\]