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
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![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](/uploads/image/z/5256323-35-3.jpg?t=1.%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8D%E5%9B%9B%E9%A1%B9%E5%92%8C%E7%AD%89%E4%BA%8E4%2C%E8%AE%BE%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94n%E2%89%A52%E6%97%B6%2Can%3D1%2F2%EF%BC%88%E6%A0%B9%E5%8F%B7Sn%2B%E6%A0%B9%E5%8F%B7Sn-1%29%2C%E6%B1%82S102.%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E8%A7%92A.B.C%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%E5%88%86%E5%88%AB%E6%98%AFa.b.c%2CtanA%3D1%2F2%2CcosB%3D3%E5%80%8D%E6%A0%B9%E5%8F%B710%2F10%2C%E2%91%A0%E6%B1%82tanC%E7%9A%84%E5%80%BC%EF%BC%9B%E2%91%A1%E8%8B%A5%E2%96%B3ABC%E7%9A%84%E6%9C%80%E9%95%BF%E8%BE%B9%E6%98%AF1%2C%E6%B1%82%E6%9C%80%E7%9F%AD%E8%BE%B9)
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
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.
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
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
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
1.s10=49~2<1>tanC=1~<2>最小边是根号5/5