s 1 = 2+(2+3)/ 1+3 = 2+5+3 = 10

S2 = 2+(2+5)/2+(5+3)/2+3 = 2+3.5+4+3 = 12.5

It is easy to see that every time N is operated, except the first two numbers and the last two numbers (2 and 3), the middle number n-2 has to be calculated twice, so when S(k) is known, there are:

s(k+ 1)=(2S(k)-5)/(k+ 1)+5

Among them, the numerator represents twice of S(k) minus the sum of the first and last two numbers, because they only calculate 1 times, and the last +5 means that the first and last two numbers will be added back when the sum of k+ 1 is completed.

Therefore:

S0=5

s 1 =(2 * 5-5)/ 1+5 = 10

S2=(2* 10 - 5)/2 +5=25/2

S3=(2*25/2 - 5)/3 +5=35/3

S4 =(2 * 35/3-5)/4+5 = 1 15/ 12

S5 =(2 * 1 15/ 12-5)/5+5 = 47/6

S6=(2*47/6 -5)/6 +5=6 1/9