1.(a(n+1)-an)4(an-1)+(an-1)²=0
(an-1)(4a(n+1)-4an+an-1)=0
4a(n+1)=3an+1
a(n+1)-1=(3/4)(an-1)
an-1為首項(xiàng)為1公比為3/4的等比數(shù)列
2.an-1=(3/4)^(n-1)
an=(3/4)^(n-1)+1
sn=(1-(3/4)^n)/(1-3/4)+n
sn=4-4*(3/4)^n+n≥1+n
3.bn=3(an-1)²-4(a(n+1)-1)
a(n+1)-1=3/4*(an-1)
bn=3(an-1)²-3(an-1)
bn=3((an-1)-1/2)²-3/4
bn=3((3/4)^(n-1)-1/2)²-3/4
(3/4)^(n-1)無(wú)最小值,最大為1,n=1時(shí),此時(shí)bn有最大項(xiàng),b1=0
只有(3/4)^(n-1)最接近1/2時(shí),bn有最小項(xiàng)
n=2,3/4>1/2,3/4-1/2=1/4
n=3,9/16>1/2,9/16-1/2=1/16
n=4,27/641/16
所以當(dāng)n=3時(shí),bn有最小項(xiàng),b3=3/256-3/4=-189/256