To find the sum of the first n terms, we need to find the partial sum Sn.

The general term An is given by An = 2/n^2 + 5n + 6.

To find Sn, we need to find the sum of the first n terms of An. We can do this by using partial fraction decomposition.

First, let’s factor the denominator of An: n^2 + 5n + 6 = (n + 2)(n + 3).

Now, we can write An as a sum of partial fractions:

An = A/(n + 2) + B/(n + 3).

To find the values of A and B, we can multiply both sides of the equation by the common denominator (n + 2)(n + 3):

An(n + 2)(n + 3) = A(n + 3) + B(n + 2).

Simplifying this equation, we get:

2 = A(n + 3) + B(n + 2).

Now, we can substitute values of n to solve for A and B.

Let’s start with n = -2:

2 = A(-2 + 3) + B(-2 + 2).

2 = A + 0.

A = 2.

Now, let’s try n = -3:

2 = A(-3 + 3) + B(-3 + 2).

2 = 0 + B.

B = 2.

So, we have A = 2 and B = 2.

Now, we can rewrite An as:

An = 2/(n + 2) + 2/(n + 3).

To find Sn, we can sum up the first n terms:

Sn = 2/(1 + 2) + 2/(2 + 2) + 2/(3 + 2) + … + 2/(n + 2).

Simplifying this expression, we get:

Sn = 2/3 + 2/4 + 2/5 + … + 2/(n + 2).

Now, we can find the sum of this series by taking the common denominator:

Sn = (2(n + 2) + 2(n + 1) + 2(n) + … + 2(3))/(3 * 4 * 5 * … * (n + 2)).

Simplifying this expression, we get:

Sn = (2(n + 2) + 2(n + 1) + 2(n) + … + 6)/(3 * 4 * 5 * … * (n + 2)).

Sn = (2(n + 2) + 2(n + 1) + 2(n) + … + 6)/(3 * 4 * 5 * … * (n + 2)).

Sn = (2n + 4 + 2n + 2 + 2n + … + 6)/(3 * 4 * 5 * … * (n + 2)).

Sn = (2n^2 + 8n + 6)/(3 * 4 * 5 * … * (n + 2)).

Therefore, the sum of the first n terms is given by the expression (2n^2 + 8n + 6)/(3 * 4 * 5 * … * (n + 2)).