To find the sum of the first n terms, we need to find the sum of the individual terms and add them up.

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

To find the sum of the first n terms, we need to substitute the values of n from 1 to n into the general term and add them up.

So, the sum of the first n terms (Sn) is given by:

Sn = A1 + A2 + A3 + … + An

Substituting the values of n into the general term, we get:

Sn = (2/1^2 + 5(1) + 6) + (2/2^2 + 5(2) + 6) + (2/3^2 + 5(3) + 6) + … + (2/n^2 + 5(n) + 6)

Simplifying each term, we get:

Sn = (2 + 5 + 6) + (2/4 + 10 + 6) + (2/9 + 15 + 6) + … + (2/n^2 + 5n + 6)

Sn = 13 + (2/4 + 10 + 6) + (2/9 + 15 + 6) + … + (2/n^2 + 5n + 6)

Sn = 13 + (1/2 + 5/2 + 6/2) + (1/9 + 5/3 + 6/3) + … + (1/n^2 + 5n/n + 6/n)

Sn = 13 + (1/2 + 5/2 + 6/2) + (1/9 + 5/3 + 6/3) + … + (1/n^2 + 5 + 6/n)

Sn = 13 + (12/2) + (12/3) + … + (12/n)

Sn = 13 + 6 + 4 + … + 12/n

Sn = 13 + 6(1 + 2/3 + 1/2 + … + 1/n)

Now, we need to find the sum of the reciprocals of the numbers from 1 to n.

This is a harmonic series, and its sum is given by:

1 + 1/2 + 1/3 + … + 1/n = ln(n) + γ + O(1/n)

where ln(n) is the natural logarithm of n and γ is the Euler-Mascheroni constant.

So, the sum of the first n terms is:

Sn = 13 + 6( ln(n) + γ + O(1/n) )

Therefore, the sum of the first n terms is approximately 13 + 6ln(n) + 6γ.