1, so that the derivative is equal to zero. Substitute x=- 1 and get a= 1/3, then g (x) =1/(x2+x) = (1/x)-/kloc-.

According to the meaning, s1= 0+g (1) = g (1), S2=g( 1)+g(2).

So the final S=g( 1)+g(2)+. . . +g(n)= 1-( 1/2)+( 1/2)-( 1/3)+.。 . +( 1/n)- 1/(n+ 1)= 1- 1/(n+ 1)〉20 1 1/20 12

The solution is n > 20 1 1. Note that in the end, n=n+ 1, and the answer is d.

2、5Sn=5a 1+5^2*a2+5^3*a3+...+5^n*an

Then 6sn = 5sn+sn = a1+5 (a1+a2)+5 2 * (a2+a3)+...+5 (n-1) * [a (n-1)]

Because an+a (n+1) = (1/5) n.

Therefore, the above formula can be simplified as: 6Sn = 5 * (1/5)+5 2 * (1/5) 2+...+5 (n-1) * (1/5) (n-

=n- 1+a 1+5^n*an

Therefore, the formula = (n+1)/n.