1.an=1/[根号n+根号(n+1)]
=[根号(n+1)-根号n]/
{[根号(n+1)+根号n]*[根号(n+1)-根号n]
=[根号(n+1)-根号n]/[(n+1)-n]
=根号(n+1)-根号n
由于Sn=10
Sn
=a1+a2+...+an
=an+a(n-1)+...+a1
=[根号(n+1)-根号n]+[根号n-根号(n-1)]+...+根号2-根号1]
=根号(n+1)-1
=10
则:11=根号(n+1)
121=n+1
则:n=120
2.
70q^3+70=133q+133q^2
70(q^3+1)-133q(q+1)=0
70(q+1)(q^2-q+1)-133q(q+1)=0
(q+1)(70q^2-203q+70)=0
7(q+1)(5q-2)(2q-5)=0
则:q1=-1,q2=2/5,q3=5/2