According to the meaning of the problem, we get the recurrence formula:

a _ { n+ 1 } =( 1/n)*(a_n^2+2n)

First, we can write the recurrence formula as follows:

a_{n+ 1} = a_n^2/n + 2

Then, we can compare this recursive formula with the general formula and find that this recursive formula is similar to the general formula of geometric series, namely:

a_{n+ 1} = q * a_n + c

Therefore, we can set q = 1/n and c = 2, and get the general formula:

a _ n = q^(n- 1)* a _ 1+(q^(n- 1)- 1)/(q- 1)* c

According to a 1 = 5 given in the title, the general formula is obtained:

a _ n = 1/n^(n- 1)* 5+( 1/n^(n- 1)- 1)/( 1/n- 1)* 2

As you can see, the general formula of this series is:

a _ n = 5/n^(n- 1)+2/(n- 1)

Next, let's find the first n terms and S_n of the sequence.

According to the summation formula:

S_n = a_ 1 + a_2 +...+ a_n

We can get:

S_n = 5 + 5/2 + 5/3 +...+ 5/n

It should be noted that when n is large enough, 5/n approaches 0, so we can take 5/n as a minimum and expand it by Taylor expansion, and get:

s _ n≈5 *( 1+ 1/2+ 1/3+...+ 1/n)

As we know, this formula is the form of finding harmonic series. According to the formula of harmonic series:

h _ n = 1+ 1/2+ 1/3+...+ 1/n

We have:

s n≈5 * H n

Therefore, the sum of the first n terms of the sequence s _ n ≈ 5 * h _ n