(1)S11=11a1+(1+10)10/2 *d=0
a3=a1+2d =24
a1=24-2d
11(24-2d) +55d=0
11*24-22d+55d=0
33d=-11*24
d=-8
a1=24-2d=40
所以an=a1+(n-1)d=48-8n
(2)Sn=na1+(1+n)n/2 *d=-4n^2+44n
(3) Sn=-4(n^2-11n)=-4(n-11/2)^2+11^2 顯然5< 11/23.a（n+1）=an+2根號(hào)（an+1）a(n+1)=an +2根號(hào)an +1吧a(n+1)=(根號(hào)an +1)^2a2=(根號(hào)2 +1)^2a3=(根號(hào)a2+1)^2 =(根號(hào)2+1+1)^2=(根號(hào)2+2)^2a4=(根號(hào)a3+1)^2=(根號(hào)2+2+1)^2 =（根號(hào)2+3)^2......可以猜測(cè)an=(根號(hào)2+n-1)^2用數(shù)學(xué)歸納法可以證明。