1.已知数列{an}的前四项和等于4,设前n项和为Sn,且n≥2时,an=1/2（根号Sn+根号Sn-1),求S102.在△ABC中,角A.B.C所对的边分别是a.b.c,tanA=1/2,cosB=3倍根号10/10,①求tanC的值；②若△ABC的最长边是1,求最短边的长.
问题描述：
1.已知数列{an}的前四项和等于4,设前n项和为Sn,且n≥2时,an=1/2（根号Sn+根号Sn-1),求S10
2.在△ABC中,角A.B.C所对的边分别是a.b.c,tanA=1/2,cosB=3倍根号10/10,
①求tanC的值；②若△ABC的最长边是1,求最短边的长.
答
1.s10=49～2tanC=1～最小边是根号5/5
答
Sn-Sn-1=1/2（根号Sn+根号Sn-1)
2(根号Sn-根号Sn-1)（根号Sn+根号Sn-1)=根号Sn+根号Sn-1
根号Sn=1/2根号Sn-1
根号Sn为等比数列
之后有两种方法1、依次往下求
2、先求根号S1，求得根号Sn的公式，再求Sn，最后求S10
答
1.
a[n] = S[n]-S[n-1] = 1/2 (√S[n]+√S[n-1])
==> √S[n] - √S[n-1] = 1/2
==> √S[10] - √S[4] = 1/2 * 6 = 3,√S[4]=√4=2
==> √S[10] = 5,
==> S[10] = 25
2.
tanB = sinB/cosB = 1/3
==> tan(A+B) = (1/2+1/3)/(1-1/2*1/3) = 1
==> tanC = -tan(A+B) = -1
sinA=√5/5,sinB=√10/10,sinC=√2/2
==> sinC>sinA>sinB ==> c>a>b
==> b/c = sinB/sinC = √5/5
==> b=√5/5