1/1×2+1/2×3+1/3×4+......+1/N+(N+1)

To simplify the expression given, we'll first break down how it is structured:

We have the expression:

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

Let's denote the inner sum as \( T \):

\[
T = \frac{1}{1} \cdot 2 + \frac{1}{2} \cdot 3 + \frac{1}{3} \cdot 4 + \ldots + \frac{1}{N} \cdot (N+1)
\]

We can now express \( S \) in terms of \( T \):

\[
S = T + (N+1)T = (1 + (N+1)) T = (N+2) T
\]

Now we need to compute \( T \):

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

The first sum, \( \sum_{k=1}^{N} 1 \), equals \( N \).

The second sum, \( \sum_{k=1}^{N} \frac{1}{k} \), is the \( N \)-th harmonic number, commonly denoted as \( H_N \).

Thus,

\[
T = N + H_N
\]

Substituting back into the expression for \( S \):

\[
S = (N+2)(N + H_N) = (N+2)N + (N+2)H_N = N^2 + 2N + (N+2)H_N
\]

So the final expression for \( S \) is:

\[
S = N^2 + 2N + (N+2)H_N
\]

This gives you the complete simplified expression.